Question 13 Marks
Find the value of k for which the following systems of equations has unique solution:
$x - ky = 2, 3x + 2y + 5 = 0$
Answer$x - ky = 2, 3x + 2y + 5 = 0$
$x - ky - 2 = 0, 3x + 2y + 5 = 0$
We know that, the system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
Has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{1}{3}\neq\frac{\text{k}}{2}$
$\Rightarrow\ \text{k}\neq\frac{-2}{3}$
View full question & answer→Question 23 Marks
Find the value of k for which the following systems of equations has unique solution:
$4x + ky + 8 = 0, x + y + 1 = 0$
Answer$4x + ky + 8 = 0, x + y + 1 = 0$
We know that, the system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
$\Rightarrow\ \frac{4}{1}\neq\frac{\text{k}}{1}$
$\Rightarrow\ \text{k}\neq4.$
View full question & answer→Question 33 Marks
Show that the system of equations:
$2x - 3y = 5, 6x - 9y = 15$
has an infinite number of solutions.
Answer$2x - 3y - 5 = 0,$
$6x - 9y - 15 = 0$
These equations are of the form
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Where $a_1 = 2, b_1 = -3, c_1 = -5$
$a_2= 6, b_2 = -9, c_2 = -15$
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{6}=\frac{1}{3},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-3}{-9}=\frac{1}{3}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{-5}{-15}=\frac{1}{3}$
Thus, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Hence the given system of equations has infinitely many solutions.
View full question & answer→Question 43 Marks
Solve for x and y:
$\text{4x}-\text{3y}=8,$
$\text{6x}-\text{y}=\frac{29}{3}$
AnswerThe given equations are: $\text{4x}-\text{3y}=8\ \dots(1)$ $\text{6x}-\text{y}=\frac{29}{3}\ \dots(2)$ Multiply (1) by 2 and 2 by 3 $\text{4x}-\text{3y}=8\ \dots(3)$ $\text{18x}-\text{3y}=29\ \dots(4)$Subtracting (3) from (4), we get
$\text{14x}=21$ $\Rightarrow\text{x}=\frac{21}{14}=\frac{3}{2}$Substitution $\text{x}=\frac{3}{2}$ in (1), we get
$2\times\frac{3}{2}-\text{3y}=8$ $\Rightarrow6-\text{3y}=8$ $-\text{3y}=2$ $\Rightarrow\text{y}=\frac{-2}{3}$$\therefore$ Solution is $\text{x}=\frac{3}{2}$ and $\text{y}=\frac{-2}{3}$
View full question & answer→Question 53 Marks
Solve for x and y:
$\text{x}-\text{y}=3$
$\frac{\text{x}}{3}+\frac{\text{y}}{2}=6$
AnswerThe given equation are:
$\text{x}-\text{y}=3\dots(\text{i})$
$\frac{\text{x}}{3}+\frac{\text{y}}{2}=6\dots(\text{ii})$
Multiply (i) by $\frac{1}2{}$ and add it to (ii).
$\frac{\text{x}}{2}-\frac{\text{y}}{2}=\frac{3}{2}$ and $\frac{\text{x}}{3}+\frac{\text{y}}{2}=6$
$\Rightarrow\frac{\text{x}}2{}+\frac{\text{x}}{3} =\frac{3}{2}+6$
$\Rightarrow\frac{\text{5x}}6{}=\frac{15}2{}$
$\Rightarrow\text{x}=9$
Substituting in (i), we get y = 6
So, x = 9 and y = 6
View full question & answer→Question 63 Marks
Solve for x and y:
0.3x + 0.5y = 0.5,
0.5x + 0.7y = 0.74
AnswerThe given equations are: 0.3x + 0.5y = 0.5 ...(i) 0.5x + 0.7y = 0.74 ...(ii)Multiply (i) by 5 and (ii) by 3 and subtract (ii) from (i).
⇒ 2.5y - 2.1y = 2.5 - 2.22
⇒ 0.4y = 0.28
⇒ y = 0.7
Substitute y = 0.7 in (i), we get
⇒ 0.3x + 0.5(0.7) = 0.5
⇒ x = 0.5
So, x = 0.5 and y = 0.7
View full question & answer→Question 73 Marks
Find the value of k for which the following systems of equations has unique solution:
$kx + 3y = (k - 3), 12x + ky = k$
Answer$kx + 3y - (k - 3) = 0$
$12x + ky - k = 0$
These equations are of the form
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Where $a_1 = k, b_1 = 3, c_1 = -(k - 3)$
$a_2= 12, b_2 = k, c_2 = -k$
For unique solution, we have
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\frac{\text{k}}{12}\neq\frac{3}{\text{k}}$
$\text{k}^2\neq36$
$\Rightarrow\text{k}\neq\pm6$
Thus, for all real value of k other than $\pm6,$ the given system of equations will have a unique solution.
View full question & answer→Question 83 Marks
Very-Short and Short-Answer Questions:
Write the number of solutions of the following pair of linear equations:
$x + 3y - 4 = 0, 2x + 6y - 7 = 0$
AnswerThe given pair of linear equation are:
$x + 3y - 4 = 0 ...(i)$
$2x + 6y - 7 = 0 ...(ii)$
Which is of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0,$
where $a_1 = 1, b_1 = 3, c_1 = -4, a_2 = 2, b_2 = 6$ and $c_2 = 7$
Now
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-4}{-7}=\frac{4}{7}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
Thus, the pair of the given linear equations has no solution.
View full question & answer→Question 93 Marks
Solve for x and y:
$\text{2x}+\text{5y}=\frac{8}{3},$
$3\text{x}-\text{2y}=\frac{5}{6}$
AnswerThe given equations are: $\text{2x}+\text{5y}=\frac{8}{3}\ \dots(1)$ $\text{3x}-\text{2y}=\frac{5}{6}\ \dots(2)$ Multiply (1) by 2 and (2) by 5 $\text{4x}+\text{10y}=\frac{16}{3}\ \dots(3)$ $\text{15x}-\text{10y}=\frac{25}{6}\ \dots(4)$Adding (3) from (4), we get
$19\text{x}=\frac{57}6{}$ $\Rightarrow\text{x}=\frac{57}{6\times19}=\frac{1}{2}$Substitution $\text{x}=\frac{1}{2}$ in (3), we get
$2\times\frac{1}{2}+\text{10y}=\frac{16}{3}$ $\text{10y}=\frac{16}{3}-2$ $\Rightarrow\text{10y}=\frac{10}{3}$ $\text{y}=\frac{10}{3\times10}=\frac{1}{3}$$\therefore$ Solution is $\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}{3}$
View full question & answer→Question 103 Marks
Find the value of $k$ for which the following systems of linear equations has an infinite number of solutions:
$(k - 1)x - y = 5,$
$(k + 1)x + (1 - k)y = (3k + 1)$
Answer$(k - 1)x - y = 5, (k + 1)x + (1 - k)y = (3k + 1)$ These are of the form $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$ Where $a_1 = 5, b_1 = 2, c_1 = -2k a_2= 2(k + 1), b_2 = k, c_2 = -(3k + 4)$ For infinite number of solutions, we must have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\frac{\text{k}-1}{\text{k}+1}=\frac{-1}{-(\text{k}-1)}=\frac{-5}{-(\text{3k}+1)}$
$\frac{\text{k}-1}{\text{k}+1}=\frac{1}{(\text{k}-1)}=\frac{5}{(\text{3k}+1)}$
Case I:
$\frac{\text{k}-1}{\text{k}+1}=\frac{1}{(\text{k}-1)}$
$(k - 1)^2 = k + 1$
$\Rightarrow k^2 + 1 - 2k = k + 1$
$\Rightarrow k^2 + 1 - 1 - 2k - k = 0$
$\Rightarrow k^2 = 3k \Rightarrow k = 3$
Case II:
$\frac{1}{(\text{k}-1)}=\frac{5}{(\text{3k}+1)}$
$(3k + 1) = 5(k - 1)$
$\Rightarrow 3k + 1 = 5k - 5$
$\Rightarrow -2k = -6$
$\Rightarrow k = 3$
Case III:
$\frac{\text{k}-1}{\text{k}+1}=\frac{5}{(\text{3k}+1)}$
$\Rightarrow (k - 1)(3k + 1) = 5(k + 1)$
$\Rightarrow 3k^2 + k - 3k - 1 = 5k + 5$
$\Rightarrow 3k^2- 2k - 5k - 1 - 5 = 0$
$\Rightarrow 3k^2 - 7k - 6 = 0$
$\Rightarrow 3k^2 - 9k + 2k - 6 = 0$
$\Rightarrow 3k(k - 3) + 2(k - 3) = 0$
$\Rightarrow (3k + 2)(k - 3) = 0$
$\Rightarrow 3k = -2$ or $k = 3$
$\Rightarrow\text{k}=\frac{-2}{3}$ or $k = 3$
Thus, for $k = 3,$ is the common for which there are infinitely many solutions of the given system of equations.
View full question & answer→Question 113 Marks
Find the value of a and b for which the following systems of linear equations has an infinite number of solutions:
$(a - 1)x + 3y = 2,$
$6x + (1 - 2b)y = 6$
Answer$(a - 1)x + 3y - 2 = 0,$
$6x + (1 - 2b)y - 6 = 0$
These are of the form
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Where $a_1 = (a - 1), b_1 = 3, c_1 = -2$
$a_2= 6, b_2 = (1 - 2b), c_2 = -6$
For infinite number of solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\frac{(\text{a}-1)}{6}=\frac{3}{(1-\text{2b})}=\frac{-2}{-6}$
$\Rightarrow\frac{(\text{a}-1)}{6}=\frac{3}{(1-\text{2b})}=\frac{1}{3}$
$\Rightarrow\frac{(\text{a}-1)}{6}=\frac{1}{3}$ and $\frac{3}{(1-\text{2b})}=\frac{1}{3}$
$\Rightarrow\text{3a}-3=6$ and $9=1-\text{2b}$
$\Rightarrow\text{3a}=6+3$ and $\text{2b}=1-9$
$\text{3a}=9$
$\Rightarrow\text{a}=\frac{9}{3}=3$ and $\text{2b}=-8$
$\text{b}=\frac{-8}{2}=-4$
Hence $a = 3$ and $b = -4$
View full question & answer→Question 123 Marks
Solve for x and y:
$\text{2x}+\text{3y}+1=0,$
$\frac{7-\text{4x}}{3}=\text{y}$
AnswerThe given equations are: 7 - 4x = 3y -4x - 3y = -7 4x + 3y = 7 ...(1) 2x + 3y = -1 ...(2)Subtracting (2) from (1), we get
2x = 8
$\therefore$ x = 4
Substitution x = 4 in (1), we get
4 × 4 + 3y = 7
⇒ 3y = 7 - 16
⇒ 3y = -9
⇒ y = -3
$\therefore$ Solution is x = 4 and y = -3
View full question & answer→Question 133 Marks
Very-Short and Short-Answer Questions:Write the number of solutions of the following pair of linear equations:
$x + 2y - 8 = 0,$
$2x + 4y = 16$
AnswerThe given equations are:
$x + 2y - 8 = 0 ...(i)$
$2x + 4y - 16 = 0 ...(ii)$
which is of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, where
$a_1 = 1, b_1 = 2, c_1 = -8, a_2 = 2, b_2 = 4$ and $c_2 = -18$
Now,
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-16}=\frac{1}{2}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Thus, the pair of linear equations are coincident and therefore has infinitely many solutions.
View full question & answer→Question 143 Marks
Find the value of a and b for which the following systems of linear equations has an infinite number of solutions:
$2x + 3y = 7,$
$2ax + (a + b)y = 28$
AnswerWe know that,
if in a system of linear equations
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
has infinite number of solutions, then $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Given that,
$2x + 3y = 7, 2ax + (a + b)y = 28$
have an infinite number of solutions.
$2x + 3y - 7 = 0$ and $2ax + (a + b)y - 28 = 0$
Since the pair of lines have an infinite number of solutions,
So, $\frac{2}{\text{2a}}=\frac{3}{\text{a}+\text{b}}=\frac{-7}{-28}$
$\Rightarrow\frac{1}{\text{a}}=\frac{3}{\text{a}+\text{b}}=\frac{7}{28}$
$a = 4$ and
$\Rightarrow a + b = 3a$
$\Rightarrow 4 + b = 12$
$\Rightarrow b = 8$
Hence, $a = 4$ and $b = 8$
View full question & answer→Question 153 Marks
Find the value of k for which the following systems of equations has no solution:
$kx + 3y = k - 3,$
$12x + ky = k$
Answer$kx + 3y = k - 3, 12x + ky = k$
$\Rightarrow kx + 3y - (k - 3) = 0, 12x + ky - k = 0$
We know that, the system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
has a no solutions, if $\frac{\text{a}_1}
{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
That is, $\frac{\text{k}}{\text{12}}=\frac{\text{3}}{\text{k}}\neq\frac{-(\text{k}-3)}{-\text{k}}$
So, $\text{k}^2=36$ $\Rightarrow\text{k}=\pm6$ On solving $\frac{3}{\text{k}}\neq\frac{3-\text{k}}{-\text{k}},$
we get$\frac{3}{1}\neq\frac{3-\text{k}}{-1}$
$-3\neq3-\text{k}$
$\Rightarrow\text{k}\neq6$ So, $k = -6$
View full question & answer→Question 163 Marks
Find the value of a and b for which the following systems of linear equations has an infinite number of solutions:
$(2a - 1)x + 3y = 5,$
$3x + (b - 1)y = 2$
Answer$(2a - 1)x + 3y - 5 = 0,$
$3x + (b - 1)y - 2 = 0$
These equations are of the form
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Where $a_1 = (2a - 1), b_1 = 3, c_1 = -5$
$a_2= 3, b_2 = (b - 1), c_2 = -2$
These holds only when
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\frac{(\text{2a}-1)}{3}=\frac{3}{(\text{b}-1)}=\frac{-5}{-2}$
$\Rightarrow\frac{(\text{2a}-1)}{3}=\frac{3}{(\text{b}-1)}=\frac{5}{2}$
$\Rightarrow\frac{(\text{2a}-1)}{3}=\frac{5}{2}$ and $\frac{3}{(\text{b}-1)}=\frac{5}{2}$
$\Rightarrow\text{4a}-2=15$ and $5(\text{b}-1)=6$
$\Rightarrow\text{4a}=17$ and $5\text{b}-5=6$
$\Rightarrow\text{a}=\frac{17}{4}$ and
$\text{5b}=11$
$\Rightarrow\text{b}=\frac{11}{5}$
Hence $\text{a}=\frac{17}{4}$ and $\text{b}=\frac{11}{5}$
View full question & answer→Question 173 Marks
Solve the following systems of equations has unique solution and solve it:
$2x - 3y = 17, 4x + y = 13$
Answer$2x - 3y = 17, 4x + y = 13$
$2x - 3y - 17 = 0, 4x + y - 13 = 0$
We know that, the system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
Has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{1}=-3$
Clearly, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, the system has a unique solution.
$2x - 3y = 17 ...(1)$
$4x + y = 13 ...(2)$
Multiply (2) by 3 and add to (1).
$14x = 56$ ? $x = 4$
Substituting $x = 4$ in (1), we get $y = -3.$
So, the solution is $x = 4, y = -3.$
View full question & answer→Question 183 Marks
Solve the following systems of equations has unique solution and solve it:
$3x + 5y = 12, 5x + 3y = 4$
Answer$3x + 5y - 12 = 0, 5x + 3y - 4 = 0$
$a_1 = 3, b_1 = 5, c_1 = -12$
$a_2 = 5, b_2 = 3, c_2 = -4$
Thus, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}=\Big(\frac{3}{5}\neq\frac{5}{3}\Big)$
Hence, the given system of equations has a unique solution.
The given equations are:
$3x + 5y - 12 = 0 ...(1)$
$5x + 3y - 4 = 0 ...(2)$
Multiplying $(1)$ by $3$ and $(2)$ $5$, we get
$9x + 15y = 36 ...(3)$
$25x + 15y = 20 ...(4)$
Subtracting $(3)$ from $(4)$, we get
$16x = -16$
$\Rightarrow\ \text{x}=\frac{-16}{16}=-1$
Putting $x = -1$, in (3), we get
$9 × (-1) + 15y = 36$
$-9 + 15y = 36$
$15y = 36 + 9$
$\Rightarrow\ \text{y}=\frac{45}{15}=3$
$\therefore$ The solution is $x = -1, y = 3$
View full question & answer→Question 193 Marks
Very-Short and Short-Answer Questions:
Write the number of solutions of the following pair of linear equations:
$2x + 3y = 7$
$(k - 1)x + (k + 2)y = 3k$
AnswerThe given equations are:
$2x + 3y - 7 = 0 ...(i)$
$(k - 1)x + (k + 2)y - 3k = 0 ...(ii)$
which is of the form $a_1x + b_1y + c_1 = 0 $and $a_2x + b_2y + c_2 = 0$, where
$a_1 = 2, b_1 = 3, c_1 = -7, a_2 = k - 1, b_2 = k + 2$ and $c_2 = -3k$
For the given pair of linear equations to have infinitely many solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{\text{k}-1}=\frac{3}{\text{k}+2}=\frac{-7}{-\text{3k}}$
$\Rightarrow\frac{2}{\text{k}-1}=\frac{3}{\text{k}+2},$ $=\frac{3}{\text{k}+2}=\frac{-7}{-\text{3k}}$ and $=\frac{2}{\text{k}-1}=\frac{-7}{-\text{3k}}$
$\Rightarrow 2(k + 2) = 3(k - 1), 9k = 7k + 14$ and $6k = 7k - 7$
$\Rightarrow k = 7, k = 7$ and $k = 7$
Hence, $k = 7$
View full question & answer→Question 203 Marks
Find the value of k for which the following systems of linear equations has an infinite number of solutions:
$5x + 2y = 2k,$
$2(k + 1)x + ky = (3k + 4).$
Answer$5x + 2y - 2k = 0, $
$2(k + 1)x + ky - (3k + 4) = 0.$
These are of the form $a_1x + b_1y + c_1 = 0, $
$a_2x + b_2y + c_2 = 0. $
Where $a_1 = 5, b_1 = 2, c_1 = -2k $
$a_2= 2(k + 1), b_2 = k, c_2 = -(3k + 4)$
For infinite number of solutions,
we must have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
This hold only when $\frac{\text{5}}{2(\text{k}+1)}=\frac{2}{\text{k}}=\frac{-\text{2k}}{-(3\text{k}+4)}$
$\Rightarrow\frac{\text{5}}{2(\text{k}+1)}=\frac{2}{\text{k}}=\frac{\text{2k}}{(3\text{k}+4)}$
Now the following cases arises
Case: I
$\frac{\text{5}}{2(\text{k}+1)}=\frac{2}{\text{k}}$
$\Rightarrow 5k = 4(k + 1) $
$\Rightarrow 5k = 4k + 4 $
$\Rightarrow k = 4$
Case: II
$\frac{\text{2}}{\text{k}}=\frac{\text{2k}}{(\text{3k}+4)}$
$\Rightarrow 2(3k + 4) = 2k^2 $
$\Rightarrow 6k + 8 = 2k^2 $
$\Rightarrow 2k^2 - 6k - 8 = 0$
$ \Rightarrow 2(k^2 - 3k - 4) = 0 $
$\Rightarrow k^2- 3k - 4 = 0 $
$\Rightarrow k^2 - 4k + k - 4 = 0 $
$\Rightarrow k(k - 4) + 1(k - 4) = 0$
$ \Rightarrow (k - 4)(k + 1) = 0 $
$\Rightarrow (k - 4) = 0$ or $k + 1= 0$
$ \Rightarrow k = 4$ or $k = -1$
Case: III
$\frac{\text{5}}{2(\text{k}+1)}=\frac{\text{2k}}{(\text{3k}+4)}$
$\Rightarrow 15k + 20 = 4k^2 = 4k $
$\Rightarrow 4k^2 + 4k - 15k - 20 = 0 $
$\Rightarrow 4k^2- 11k - 20 = 0 $
$\Rightarrow 4k^2- 16k + 5k - 20 = 0 $
$\Rightarrow 4k(k - 4) + 5(k - 4) = 0$
$ \Rightarrow (k - 4)(4k + 5) = 0 $
$\Rightarrow k = 4$ or $\text{k}=\frac{-5}{4}$
Thus, for $k = 2,$ is the common for which there are infinitely many solutions of the given system of equations.
View full question & answer→Question 213 Marks
Solve the following systems of equations has unique solution and solve it:
$\frac{\text{x}}{3}+\frac{\text{y}}2{}=3,\ \text{x}-\text{2y}=2$
Answer$\frac{\text{x}}{3}+\frac{\text{y}}{2}=3$
$\Rightarrow\ \frac{2\text{x}+3\text{y}}{6}=3$
$2x + 3y - 18 = 0 ...(1)$
$x - 2y - 2 = 0 ...(2)$
$a_1= 2, b_1 = 3, c_1 = -18$
$a_2 = 1, b_2 = -2, c_2 = -2$
Thus, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}\Rightarrow\ \frac{2}{1}\neq\frac{3}{-2}$
Hence, the given system of equations has unique solution.
Then given equations are
$2x + 3y = 18 ...(1)$
$x - 2y = 2 ...(2)$
Multiplying (1) by $2$ and (2) by $3$
$4x + 6y = 36 ...(3)$
$3x - 6y = 6 ...(4)$
Adding (3) and (4) we get
$7x = 42$
$\Rightarrow x = 6$
Putting $x = 6$ in (1), we get
$2 \times 6 + 3y = 18$
$\Rightarrow 3y = 18 - 12$
$\Rightarrow 3y = 6$
$\Rightarrow\text{y}=\frac{6}{3}=2$
$\therefore$ Solution is $x = 6, y = 2.$
View full question & answer→Question 223 Marks
Find the value of k for which the following systems of linear equations has an infinite number of solutions:
$kx + 3y = (2k + 1),$
$2(k + 1)x + 9y = (7k + 1).$
Answer$kx + 3y - (2k + 1) = 0,$
$2(k + 1)x + 9y - (7k + 1) = 0.$
These are of the form
$a_1x + b_1y + c_1 = 0,$
$a_2x + b_2y + c_2 = 0.$
Where $a_1 = k, b_1 = 3, c_1 = -(2k + 1)$
$a_2= 2(k + 1), b_2 = 9, c_2 = -(7k + 1)$
For infinite number of solutions, we must have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
This hold only when $\frac{\text{k}}{2(\text{k}+1)}=\frac{3}{\text{9}}=\frac{-(\text{2k}+1)}{-(7\text{k}+1)}$
$\Rightarrow\frac{\text{k}}{2(\text{k}+1)}=\frac{1}{\text{3}}=\frac{\text{2k}+1}{7\text{k}+1}$
Now the following cases arises
Case: I
$\frac{\text{k}}{2(\text{k}+1)}=\frac{1}{\text{3}}$
$\Rightarrow 2(k + 1) = 3k $
$\Rightarrow 2k + 2 = 3k $
$\Rightarrow k = 2$
Case: II
$\frac{\text{1}}{\text{3}}=\frac{\text{2k}+1}{\text{7k}+1}$
$\Rightarrow 7k + 1 = 6k + 3 $
$\Rightarrow 7k - 6k = 3 - 1 $
$\Rightarrow k = 2$
Case: III
$\frac{\text{k}}{2(\text{k}+1)}=\frac{\text{2k}+1}{\text{7k}+1}$
$\Rightarrow k(7k + 1) = 2(2k + 1)(k + 1) $
$\Rightarrow 7k^2 + k = 2(2k^2 + 2k + k + 1) $
$\Rightarrow 7k^2+ k = 2(2k^2 + 3k + 1)$
$ \Rightarrow 7k^2+ k = 4k^2+ 6k + 2 $
$\Rightarrow 7k^2 - 4k^2 + k - 6k - 2 = 0 $
$\Rightarrow 3k^2 - 5k - 2 = 0 $
$\Rightarrow 3k^2 - (6k - 1k) - 2 = 0 $
$\Rightarrow 3k(k - 2) + 1(k - 2) = 0 $
$\Rightarrow (k - 2)(3k + 1) = 0$
$ \Rightarrow k = 2$ or $\text{k}=\frac{-1}{3}$
Thus, for $k = 2,$ is the common for which there are infinitely many solutions of the given system of equations.
View full question & answer→Question 233 Marks
Find the value of k for which the following systems of linear equations has an infinite number of solutions:
$2x + 3y = 7,$
$(k - 1)x + (k + 2)y = 3k.$
Answer$2x + 3y - 7 = 0, $
$(k - 1)x + (k + 2)y - 3k = 0$
These are of the form $a_1x + b_1y + c_1 = 0,$
$a_2x + b_2y + c_2 = 0.$
Where $a_1 = 2, b_1 = 3, c_1 = -7 $
$a_2= (k - 1), b_2 = (k + 2), c_2 = -3k$
For infinite number of solutions we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
This hold only when $\frac{2}{\text{k}-1}=\frac{3}{\text{k}+2}=\frac{-7}{-3\text{k}}$
$\frac{2}{\text{k}-1}=\frac{3}{\text{k}+2}=\frac{7}{3\text{k}}$
Now the following cases arises
Case: I
$\frac{2}{\text{k}-1}=\frac{3}{\text{k}+2}$
$\Rightarrow 2(k + 2) = 3(k - 1) $
$\Rightarrow 2k + 4 = 3k - 3 $
$\Rightarrow k = 7$
Case: II
$\frac{3}{\text{k}+2}=\frac{7}{\text{3k}}$
$\Rightarrow 7(k + 2) = 9k $
$\Rightarrow 7k + 14 = 9k $
$\Rightarrow k = 7$
Case: III
$\frac{2}{\text{k}-1}=\frac{7}{\text{3k}}$
$\Rightarrow 7k - 7 = 6k $
$\Rightarrow k = 7$
For $k = 7$, there are infinitely many solutions of the given system of equations.
View full question & answer→Question 243 Marks
Solve for x and y:
2x - y + 3 = 0,
3x - 7y + 10 = 0
Answer2x - y + 3 = 0
⇒ 2x - y = - 3 ...(i)
3x - 7y + 10 = 0
⇒ 3x - 7y = 10 ...(ii)
Multiply (i) by -7 and add it to (ii).
-14x + 7y = 21 and 3x - 7y = -10
⇒ -11x = 11
⇒ x = -1
Substituting x = 1 in (i), we get y = 1
So, x = -1 and y = 1
View full question & answer→Question 253 Marks
Very-Short and Short-Answer Questions:Write the number of solutions of the following pair of linear equations:
$x + 2y - 8 = 0,$
$2x + 4y = 16$
AnswerThe given equations are:
$x + 2y - 8 = 0 ...(i)$
$2x + 4y - 16 = 0 ...(ii)$
which is of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, where
$a_1 = 1, b_1 = 2, c_1 = -8, a_2 = 2, b_2 = 4$ and $c_2 = -18$
Now,
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-16}=\frac{1}{2}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Thus, the pair of linear equations are coincident and therefore has infinitely many solutions.
View full question & answer→Question 263 Marks
Find the value of a and b for which the following systems of linear equations has an infinite number of solutions:
$2x + 3y = 7,$
$(a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1$
AnswerWe know that,
if in a system of linear equations
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
has infinite number of solutions, then $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Given that,
$2x + 3y = 7, (a + b + 1)x + (a + 2b + 2)y = 4(a + b) - 1 = 0$
Since the pair of lines have an infinite number of solutions,
So, $\frac{2}{\text{a}+\text{b}+1}=\frac{3}{\text{a}+\text{2b}+2}=\frac{-7}{-4(\text{a}+\text{b})-1}$
$\Rightarrow\frac{2}{\text{a}+\text{b}+1}=\frac{3}{\text{a}+\text{2b}+2}=\frac{7}{4(\text{a}+\text{b})+1}$
$2(a + 2b + 2) = 3(a + b + 1)$ and $7(a + 2b + 2) = 3(4(a + b) + 1)$
$\Rightarrow 2a + 4b + 4 = 3a + 3b + 3$ and $7a + 14b + 14 = 12a + 12b + 3$
$\Rightarrow a - b = 1 ...(i)$ and $5a - 2b = 11 ...(ii)$
Multiply (i) by 2 and subtract from (ii).
$\Rightarrow 2a - 2b = 2$ and $5a - 2b = 11$
$3a = 9$
$\Rightarrow a = 3$
Substituting in (i), we get $b = 2$
Hence, $a = 3$ and $b = 2$
View full question & answer→Question 273 Marks
Very-Short and Short-Answer Questions:
Write the value of k for which the system of equations $3x + ky = 0, 2x - y = 0$ has a unique solution.
AnswerThe given pair of linear equation is:
$3x + ky = 0 ...(i)$
$2x - y = 0 ...(ii)$
Which is of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,
where $a_1 = 3, b_1 = k, c_1 = 0, a_2 = 2, b_2 = -1$ and $c_2 = 0$
For the system to have a unique solution, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}$
$\frac{3}{2}\neq\frac{\text{k}}{-1}$
$\Rightarrow\text{k}\neq-\frac{3}{2}$
Hence, $\text{k}\neq-\frac{3}{2}.$
View full question & answer→Question 283 Marks
Very-Short and Short-Answer Questions: For what value of k does the following pair of linear equations have infinitely many solutions?
$10x + 5y - (k - 5) = 0$ and $20x + 10y - k = 0$
AnswerThe given pair of linear equation are:
$10x + 5y - (k - 5) = 0 ....(i)$
$20x + 10y - k = 0 ...(ii)$
Which is of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0,$
where $a_1 = 10, b_1 = 5, c_1 = -(k - 5), a_2 = 20, b_2 = 10$ and $c_2 = -k$
For the given pair of linear equation to have infinitely many solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{10}{20}=\frac{5}{10}=\frac{-(\text{k}-5)}{-\text{k}}$
$\Rightarrow\frac{1}{2}=\frac{\text{k}-5}{\text{k}}$
$\Rightarrow2\text{k}-10=\text{k}$
$\Rightarrow\text{k}=10$
Hence, $k = 10.$
View full question & answer→Question 293 Marks
Find the value of k for which the following systems of linear equations has an infinite number of solutions:
$2x + (k - 2)y = k,$
$6x + (2k - 1)y = (2k + 5).$
Answer$2x + (k - 2)y - k = 0, $
$6x + (2k - 1)y - (2k + 5) = 0.$
These are of the form
$a_1x + b_1y + c_1 = 0, $
$a_2x + b_2y + c_2 = 0.$
Where $a_1 = 2, b_1 = (k - 2),c_1 = -k $
$a_2= 6, b_2 =(2k - 1),c_2 = (2k + 5)$
For infinite number of solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
This hold only when $\frac{2}{\text{6}}=\frac{\text{k}-2}{\text{2k}-1}=\frac{-\text{k}}{-(2\text{k}+5)}$
$\frac{1}{\text{3}}=\frac{\text{k}-2}{\text{2k}-1}=\frac{\text{k}}{2\text{k}+5}$
Now the following cases arises
Case: I
$\frac{1}{\text{3}}=\frac{\text{k}-2}{2\text{k}-1}$
$\Rightarrow 2k - 1 = 3k - 6$
$ \Rightarrow k = 5$
Case: II
$\frac{\text{k}-2}{\text{2k}-1}=\frac{\text{k}}{\text{2k}+5}$
$\Rightarrow (k - 2)(2k + 5) = k(2k - 1) $
$\Rightarrow k + k = 10 $
$\Rightarrow 2k = 10$
$\Rightarrow\text{k}=\frac{10}{2}=5$
Case: III
$\frac{1}{\text{3}}=\frac{\text{k}}{\text{2k}+5}$
$\Rightarrow 2k + 5 = 3k $
$\Rightarrow 3k - 2k = 5 $
$\Rightarrow k = 5$
Thus, For $k = 5$ there are infinitely many solutions of the given system of equations.
View full question & answer→Question 303 Marks
Find the value of k for which the following systems of equations has no solution:
$3x - y - 5 = 0,$
$6x - 2y + k = 0$ $(\text{k}\neq0).$
Answer$3x - y - 5 = 0, 6x - 2y + k = 0$
We know that,
the system of linear equations
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
has a no solutions, if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
That is, $\frac{\text{3}}{\text{6}}=\frac{\text{-1}}{\text{-2}}\neq\frac{-5}{\text{k}}$
So, $-\text{k}\neq10$
$\Rightarrow\text{k}\neq-10$
View full question & answer→Question 313 Marks
Solve for x and y:
3x - 5y - 19 = 0,
-7x + 3y + 1 = 0
AnswerThe given equation are: 3x - 5y - 19 = 0 ...(1) -7x + 3y + 1 = 0 ...(2) On multiplying (1) by 3 and (2) by 5, we get: 9x - 15y = 57 ...(3) -35x + 15y = -5 ...(4) On adding (3) and (4), we get: -26x = 52⇒ x = -2
On substituting the value of x = -2 in (1), we get: 3 × (-2) - 5y = 19⇒ -6 - 5y = 19
-5y = 19 + 6
⇒ -5y = 25 y = -5 $\therefore$ Solution is x = -2 and y = -5
View full question & answer→Question 323 Marks
Solve for x and y:
$\frac{\text{x}}{2}-\frac{\text{y}}2{}=6,$
$\frac{\text{x}}{7}+\frac{\text{y}}{3}=5$
Answer$\frac{\text{x}}{2}-\frac{\text{y}}2{}=6\ \dots(\text{i})$
$\frac{\text{x}}{7}+\frac{\text{y}}{3}=5\ \dots(\text{ii})$
Multiply (ii) by $\frac{1}{3}$ and add it to (i).
$\frac{\text{x}}{21}+\frac{\text{y}}9{}=\frac{5}{3}$ and $\frac{\text{x}}{2}-\frac{\text{y}}{9}=6$
$\Rightarrow\frac{\text{x}}{21}+\frac{\text{x}}{2}=\frac{5}{3}+6$
$\Rightarrow\frac{23\text{x}}{42}=\frac{23}{3}$
$\Rightarrow\text{x}=14$
Substituting x = 14 in (ii), we get y = 9
So, x = 14 and y = 9
View full question & answer→Question 333 Marks
Find the value of a and b for which the following systems of linear equations has an infinite number of solutions:
$2x - 3y = 7,$
$(a + b)x - (a + b - 3)y = 4a + b$
Answer$2x - 3y - 7 = 0,$
$(a + b)x - (a + b - 3)y - 4a + b = 0$
These equations are of the form
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Where $a_1 = 2, b_1 = -3, c_1 = -7$
$a_2= (a + b), b_2 = (a + b - 3), c_2 = -(4a + b)$
For infinite number of solutions
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\frac{2}{\text{a}+\text{b}}=\frac{-3}{-(\text{a}+\text{b}-3)}=\frac{-7}{-(\text{4a}+\text{b})}$
$\Rightarrow\frac{2}{\text{a}+\text{b}}=\frac{3}{(\text{a}+\text{b}-3)}=\frac{7}{(\text{4a}+\text{b})}$
$\Rightarrow\frac{2}{\text{a}+\text{b}}=\frac{7}{(\text{4a}+\text{b})}$ or $\frac{3}{(\text{a}+\text{b}-3)}=\frac{7}{(\text{4a}+\text{b})}$
$8a + 2b = 7a + 7b$ and $12a + 3b = 7a + 7b - 21$
$a - 5b = 0 ...(1)$
$5a - 4b = -21 ...(2)$
Putting $a = 5b in (2)$, we get
$5 × 5b - 4b = -21$
$25b - 4b = -21$
$21b = -21$
$\text{b}=\frac{-21}{21}=-1$
Putting $b = -1$ in (1), we get
$a - 5 × -1 = 0$
$a + 5 = 0$
$a = -5$
Thus, $a = -5, b = -1$
View full question & answer→Question 343 Marks
Find the value of k for which the following systems of equations has no solution:
$kx + 3y = 3,$
$12x + ky = 6$
Answer$kx + 3y = 3, 12x + ky = 6$
$\Rightarrow kx + 3y - 3 = 0, 12x + ky - 6 = 0$
We know that,
the system of linear equations
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
has a no solutions, if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\text{k}}{\text{12}}=\frac{\text{3}}{\text{k}}\neq\frac{-3}{-6}$
$\Rightarrow\text{k}^2\neq36$
$\Rightarrow\text{k}\neq6$
But, $\frac{3}{\text{k}}\neq\frac{-3}{-6}$
$\Rightarrow-3\text{k}\neq-18$
$\Rightarrow\text{k}\neq6$
So, $k = -6$
View full question & answer→Question 353 Marks
Solve for x and y:
2x - 3y = 13,
7x - 2y = 20
AnswerThe given equation are: 2x - 3y = 13 ...(1) 7x - 2y = 20 ...(2) On multiplying (1) by 2 and (2) by 3, we get: 4x - 6y = 26 ...(3) 21x - 6y = 60 ...(4) On subtracting (3) and (4), we get: 17x = 34 x = 2 On substituting the value of x = 2 in (1), we get: 2 × 2 - 3y = 13⇒ 4 - 3y = 13
-3y = 13 - 4
⇒ -3y = 9 y = -3 $\therefore$ Solution is x = 2 and y = -3
View full question & answer→Question 363 Marks
Solve for x and y:
2x + 3y = 0,
3x + 4y = 5
AnswerThe given equation are:
2x + 3y = 0 ...(1)
3x + 4y = 5 ...(2)
On multiplying (1) by 4 and (2) by 3, we get:
8x + 12y = 0 ...(3)
9x + 12y = 15 ...(4)
On subtracting (3) and (4), we get:
x = 15
On substituting the value of x = 15 in (1), we get:
2 × 15 + 3y = 0
⇒ 3y = 0 - 30
⇒ 3y = -30 or y = -10
$\therefore$ x = 15 and y = -10
View full question & answer→Question 373 Marks
Find the value of k for which the following systems of equations:
$5x - 3y = 0, 2x + ky = 0$
Has a nonzero solution.
AnswerWe have $5x - 3y = 0 ...(1)2x + ky = 0 ...(2)$
Comparing the equations with
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
$a_1 = 5, b_1= -3, a_2 = 2, b_2 = k$
These equations have a non-zero solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}$
$\frac{\text{5}}{\text{2}}=\frac{\text{-3}}{\text{k}}$
$\Rightarrow5\text{k}=-6$
$\Rightarrow\text{k}=\frac{-6}{5}$
View full question & answer→Question 383 Marks
Find the value of k for which the following systems of linear equations has an infinite number of solutions:
$5x + 2y = 2k,$
$2(k + 1)x + ky = (3k + 4).$
Answer$5x + 2y - 2k = 0,2(k + 1)x + ky - (3k + 4) = 0.$
These are of the form
$a_1x + b_1y + c_1 = 0, $
$a_2x + b_2y + c_2 = 0.$
Where $a_1 = 5, b_1 = 2, c_1 = -2k a_2= 2(k + 1), b_2= k, c_2 = -(3k + 4)$ For infinite number of solutions,
we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
This hold only when $\frac{\text{5}}{2(\text{k}+1)}=\frac{2}{\text{k}}=\frac{-\text{2k}}{-(3\text{k}+4)}$
$\Rightarrow\frac{\text{5}}{2(\text{k}+1)}=\frac{2}{\text{k}}=\frac{\text{2k}}{(3\text{k}+4)}$
Now the following cases arises
Case: I
$\frac{\text{5}}{2(\text{k}+1)}=\frac{2}{\text{k}}$
$\Rightarrow 5k = 4(k + 1)$
$\Rightarrow 5k = 4k + 4$
$\Rightarrow k = 4$
Case: II
$\frac{\text{2}}{\text{k}}=\frac{\text{2k}}{(\text{3k}+4)}$
$\Rightarrow 2(3k + 4) = 2k^2 $
$\Rightarrow 6k + 8 = 2k^{2}$
$\Rightarrow 2k^2 - 6k - 8 = 0 $
$\Rightarrow 2(k^2 - 3k - 4) = 0 $
$\Rightarrow k^2- 3k - 4 = 0$
$\Rightarrow k^2 - 4k + k - 4 = 0 $
$\ Rightarrow k(k - 4) + 1(k - 4) = 0 $
$\Rightarrow (k - 4)(k + 1) = 0 $
$\Rightarrow (k - 4) = 0$ or $k + 1 = 0$
$ \Rightarrow k = 4$ or $k = -1$
Case: III
$\frac{\text{5}}{2(\text{k}+1)}=\frac{\text{2k}}{(\text{3k}+4)}$
$\Rightarrow 15k + 20 = 4k^2 = 4k $
$\Rightarrow 4k^2 + 4k - 15k - 20 = 0 $
$\Rightarrow 4k^2- 11k - 20 = 0 $
$\Rightarrow 4k^2- 16k + 5k^- 20 = 0 $
$\Rightarrow 4k(k - 4) + 5(k - 4) = 0 $
$\Rightarrow (k - 4)(4k + 5) = 0 $
$\Rightarrow k = 4$ or $\text{k}=\frac{-5}{4}$
Thus, for k = 2, is the common for which there are infinitely many solutions of the given system of equations.
View full question & answer→Question 393 Marks
Find the value of k for which the following systems of equations has unique solution:
$2x + 3y - 5 = 0, kx - 6y - 8 = 0$
Answer$2x + 3y - 5 = 0, kx - 6y - 8 = 0$
We know that, the system of linear equations
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
Has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{2}{\text{k}}\neq\frac{3}{-6}$
$\Rightarrow\ \text{k}\neq4.$
View full question & answer→Question 403 Marks
Find the value of k for which the following systems of equations has no solution:
$8x + 5y = 9,$
$kx + 10y = 15$
Answer$8x + 5y = 9, kx + 10y = 15$
$\Rightarrow 8x + 5y - 9 = 0, kx + 10y - 15 = 0$
We know that,
the system of linear equations
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
has a no solutions, if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{8}{\text{k}}=\frac{5}{10}\neq\frac{-9}{-15}$
$\Rightarrow\text{5k}=80$
$\Rightarrow\text{k}=16$
View full question & answer→Question 413 Marks
For what value of k does the system of equations:
$x + 2y = 5, 3x + ky + 15 = 0$
have;
- A unique solution,
- No solution?
Answer
- $x + 2y - 5 = 0,$
$3x + ky + 15 = 0$
These equations are of the form of
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Where $a_1 = 1, b_1 = 2, c_1 = -5$
$a_2= 3, b_2 = k, c_2 = 15$
For a unique solution, we must have
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ i.e., $\frac{\text{1}}{\text{3}}\neq\frac{2}{\text{k}}$
$\Rightarrow\text{k}\neq6$
Thus, for all real value of k other than 6, the given system equations will have a unique solution.
- For no solution we must have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\frac{\text{1}}{3}=\frac{2}{\text{k}}\neq\frac{-5}{15}$
$\Rightarrow\frac{1}{3}=\frac{2}{\text{k}}$ and $\frac{\text{2}}{\text{k}}\neq\frac{-5}{15}$
$\therefore\text{k}=6$
Hence, the given system of equations has no solutions when $k = 6$ View full question & answer→Question 423 Marks
For what value of k does the system of equations:
$x + 2y = 3, 5x + ky + 7 = 0$
have;
- A unique solution,
- No solution?
Also, show that there is no value of k for which the given system of equations has infinite number of solutions: Answer$x + 2y - 3 = 0,$
$5x + ky + 7 = 0$
These equations are of the form of
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Where $a_1 = 1, b_1 = 2, c_1 = -3$
$a_2= 5, b_2 = k, c_2 = 7$
- For a unique solution, we must have
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ i.e., $\frac{\text{1}}{\text{5}}\neq\frac{2}{\text{k}}$
$\Rightarrow\text{k}\neq10$
Thus, for all real value of k other than $6$, the given system equations will have a unique solution.
- For no solution we must have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\frac{\text{1}}{5}=\frac{2}{\text{k}}\neq\frac{-3}{7}$
$\Rightarrow\frac{1}{5}=\frac{2}{\text{k}}$ and $\frac{\text{2}}{\text{k}}\neq\frac{-3}{7}$
$\therefore\text{k}=10$ or $\text{k}\neq\frac{-14}{3}$
Hence, the given system of equations has no solutions when $k = 10$, $\text{k}\neq\frac{-14}{3}$
Also, For infinite number of solutions we must have:
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\frac{1}{5}=\frac{5}{\text{k}}=\frac{-3}{7}$
This is never possible since $\frac{1}{5}\neq\frac{-3}{7}$
There is no value of k for which system of equations has infinitely many solution. View full question & answer→Question 433 Marks
For what value of k does the system of equations:
$kx + 2y = 5, 3x - 4y = 10$
have;
- A unique solution,
- No solution?
Answeri.$kx + 2y - 5 = 0,$$3x - 4y - 10 = 0$
These equations are of the form
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Where $a_1 = k, b_1 = 2, c_1 = -5$
$a_2= 3, b_2 = -4, c_2 = -10$
For a unique solution, we must have
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ or $\frac{\text{k}}{\text{3}}\neq\frac{2}{-4}$
$\Rightarrow\text{k}\neq\frac{-3}{2}$
This happens when $\text{k}\neq\frac{-3}{2}$
Thus, for all real value of k other that $\frac{ -3}{2},$ the given system equations will have a unique solution.
ii.For no solution we must have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$ Now, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\frac{\text{k}}{3}=\frac{2}{-4}\neq\frac{-5}{-10}$
$\Rightarrow\frac{\text{k}}{3}=\frac{2}{-4}$ and $\frac{\text{k}}{3}\neq\frac{1}{2}$
$\text{k}=\frac{-3}{2},\ \text{k}\neq\frac{3}{2}$
Hence, the given system of equations has no solutions if $\text{k}=\frac{-3}{2}$
View full question & answer→Question 443 Marks
Solve for x and y:
x + y = 3,
4x - 3y = 26
AnswerThe given equation are:
x + y = 3 ...(1)
4x - 3y = 26 ...(2)
On multiplying (1) by 3 and (2) by 1, we get:
3x + 3y = 9 ...(3)
4x - 3y = 26 ...(4)
On adding (3) and (4), we get:
7x = 35
⇒ x = 5
On substituting the value of x = 5 in (1), we get:
x + y = 3
5 + y = 3
⇒ y = 3 - 5 = -2
$\therefore$ x = 5 and y = -2
View full question & answer→Question 453 Marks
Solve for x and y:
$\frac{\text{x}}{3}+\frac{\text{y}}{4}=11$
$\frac{\text{5x}}{6}-\frac{\text{y}}{3}=-7$
AnswerThe given equations are: $\frac{\text{x}}{3}+\frac{\text{y}}{4}=11$ $\frac{\text{5x}}{6}-\frac{\text{y}}{3}=-7$ $\frac{\text{x}}{3}+\frac{\text{y}}{4}=11$ (by taking LCM) $\frac{\text{4x}+3\text{y}}{12}=11$ $\text{4x}+\text{3y}=132\ \dots(1)$ $\frac{\text{5x}}{6}-\frac{\text{y}}{3}=-7$ (by taking LCM) $\frac{\text{5x}-\text{2y}}{6}=-7$ $\text{5x}-\text{2y}=-42\ \dots(2)$ $\text{4x}+\text{3y}=132$ $\text{5x}-\text{2y}=-42$ Multiply (1) by 2 and (2) by 3 8x + 6y = 264 ...(3) 15x - 6y = -126 ...(4)Adding (3) from (4), we get
23x = 138
⇒ x = 6
Substitution x = 6 in (1), we get
4 × 6 + 3y = 132
⇒ 3y = 132 - 24
⇒ 3y = 108
⇒ y = 36
$\therefore$ Solution is x = 6 and y = 36
View full question & answer→Question 463 Marks
Find the value of k for which the following systems of linear equations has an infinite number of solutions:
$(k - 3)x + 3y = k,$
$kx + ky = 12$
AnswerWe know that,if in a system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$ has infinite number of solutions,
then $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Given that,
$(k - 3)x + 3y = k, kx + ky = 12$
have an infinite number of solutions.
$\Rightarrow (k - 3)x + 3y - k = 0, kx + ky - 12 = 0$
Since the pair of lines have an infinite number of solutions,
So, $\frac{\text{k}-3}{\text{k}}=\frac{3}{\text{k}}=\frac{-\text{k}}{-12}$
$\Rightarrow\frac{3}{ \text{k}}=\frac{\text{k}}{12}$
$\Rightarrow\text{k}^2=36$
$\Rightarrow\text{k}=\pm6$
Solving $\frac{\text{k}-3}{\text{k}}=\frac{3}{\text{k}}$
$k - 3 = 3$ $\dots(\because\text{k}\neq0)$
$k = 6 ...$(Since we k has to satisfy all the equations, we neglect the $k = -6$)
Hence, $k = 6$
View full question & answer→Question 473 Marks
Show that the system of equations:
$\text{6x}+\text{5y}=11,\ \text{9x}+\frac{15}{2}\text{y}=21$
has no solutions.
Answer$\text{6x}+\text{5y}=11,\ \text{9x}+\frac{15}{2}\text{y}=21$
$\text{6x}+\text{5y}-11=0,$
$\text{9x}+\frac{15}{2}\text{y}-21=0$
We know that,
The system of linear equations
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
has a no solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\frac{\text{a}_1}{\text{a}_2}=\frac{6}{9}=\frac{2}{3}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{5}{\frac{15}{2}}=\frac{10}{15}=\frac{2}{3}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-11}{-21}=\frac{11}{21}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
Hence the given system of equations has no solutions.
View full question & answer→Question 483 Marks
Find the value of a and b for which the following systems of linear equations has an infinite number of solutions:
$2x + 3y = 7,$
$(a + b)x + (2a - b)y = 21$
AnswerWe know that,
if in a system of linear equations
$a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
has infinite number of solutions, then $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Given that,
$2x + 3y = 7, (a + b)x + (2a - b)y = 21$
Since the pair of lines have an infinite number of solutions,
So, $\frac{2}{\text{a}+\text{b}}=\frac{3}{\text{2a}-\text{b}}=\frac{-7}{-21}$
$\Rightarrow\frac{2}{\text{a}+\text{b}}=\frac{3}{\text{2a}-\text{b}}=\frac{1}{3}$
$a + b = 6$ and $2a - b =9$
Adding we get,
$3a = 15$
$a = 5$
Substituting in $a + b = 6$, we get $b = 1$
Hence, $a = 5$ and $b = 1$
View full question & answer→Question 493 Marks
Find the value of k for which the following systems of equations has unique solution:
$4x - 5y = k, 2x - 3y = 12$
Answer$4x - 5y - k = 0, 2x - 3y - 12 = 0$
These equations are of the form $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
Where, $a_1 = 4, b_1 = -5, c_1 = -k$
$a_2 = 2, b_2 = -3, c_2 = -12$
For unique solution, we must have
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\frac{4}{2}\neq\frac{-5}{-3}$
$2\neq\frac{5}{3}$
$\Rightarrow6\neq5$
Thus, for all real value of $k$ the given system of equations will have a unique solution.
View full question & answer→Question 503 Marks
Solve for x and y:
$\text{x}+\frac{5}{\text{y}}=\text{6},$
$\text{3x}-\frac{8}{\text{y}}=\text{5}\ (\text{y}\neq0).$
AnswerThe given equations are $\text{x}+\frac{6}{\text{y}}=6$ and $\text{3x}-\frac{8}{\text{y}}=5$ Putting $\frac{1}{\text{y}}=\text{x}$ the given equations become x + 6v = 6 ...(1) 3x - 8v = 5 ...(2)Multiplying (1) by 4 and (2) by 3, we get
4x + 24v = 24 ...(3)
9x - 24v = 15 ...(4)
Adding (3) and (4), we get
13x = 24 + 15 = 39
$\therefore\text{x}=\frac{39}{13}=3$Putting x = 3 in (1), we get
$3+\text{6v}=6$ $\therefore\text{6v}=6-3=3$ $\text{v}=\frac{3}{6}=\frac{1}{2}$ $\text{v}=\frac{1}2{}$ $\Rightarrow\frac{1}{\text{y}}=\frac{1}{2}$ $\Rightarrow\text{y}=2$$\therefore$ The solution is x = 3 and y = 2
View full question & answer→Question 513 Marks
Very-Short and Short-Answer Questions:
Solve $\frac{3}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=2$ and $\frac{9}{\text{x}+\text{y}}-\frac{4}{\text{x}-\text{y}}=1$
AnswerThe given system of equations is:
$\frac{3}{\text{x+y}}+\frac{2}{\text{x}-\text{y}}=2\ ....(\text{i})$
$\frac{9}{\text{x+y}}-\frac{4}{\text{x}-\text{y}}=1\ ...(\text{ii})$
Substituting $\frac{1}{\text{x+y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}$ in (i) and (ii), the given equations are changed to:
3v + 2v = 2 ...(iii)
9u - 4v = 1 ...(iv)
Multiplying (i) by 2 and adding it with (ii), we get
$15\text{u}=4+1$
$\Rightarrow\text{u}=\frac{1}{3}$
Multiplying (i) by 3 and subtracting (ii) from it, we get
$6\text{v}+4\text{v}=6-1$
$\Rightarrow\text{u}=\frac{5}{10}=\frac{1}{2}$
Therefore,
x + y = 3 ...(v)
x - y = 2 ....(vi)
Now, adding (v) and (vi) we have
$2\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{2}$
Substituting $\text{x}=\frac{5}{2}$ in (v), we have
$\frac{5}{2}+\text{y}=3$
$\Rightarrow\text{y}=3-\frac{5}{2}=\frac{1}{2}$
Hence, $\text{x}=\frac{5}{2}$ and $\text{y}=\frac{1}{2}.$
View full question & answer→Question 523 Marks
Find the value of k for which the following systems of equations has unique solution:
$5x - 7y - 5 = 0, 2x + ky - 1 = 0$
Answer$5x - 7y - 5 = 0, 2x + ky - 1 = 0$
We know that, the system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0.$
Has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{5}{2}\neq\frac{-7}{\text{k}}$
$\Rightarrow\ \text{k}\neq\frac{-14}{5}$
View full question & answer→Question 533 Marks
Very-Short and Short-Answer Questions:
Write the value of k for which the system of equations $3x + ky = 0, 2x - y = 0$ has a unique solution.
AnswerThe given pair of linear equation is:
$3x + ky = 0 ...(i)$
$2x - y = 0 ...(ii)$
Which is of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,
where $a_1 = 3, b_1 = k, c_1 = 0, a_2 = 2, b_2 = -1$ and $c_2 = 0$
For the system to have a unique solution, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}$
$\frac{3}{2}\neq\frac{\text{k}}{-1}$
$\Rightarrow\text{k}\neq-\frac{3}{2}$
Hence, $\text{k}\neq-\frac{3}{2}.$
View full question & answer→Question 543 Marks
Solve for x and y:
$\text{2x}-\frac{\text{3y}}4{}=3,$
$5\text{x}=2\text{y}+7$
AnswerThe given equations are: $\text{2x}-\frac{\text{3y}}{4}=3\ \dots(1)$ $\text{5x}=\text{2y}+7\ \dots(2)$ Multiply (1) by 2 and 2 by $\frac{3}{4}$ $\text{4x}-\frac{\text{3y}}{2}=6\ \dots(3)$ $\frac{15}{4}\text{x}-\frac{3}{2}\text{y}-\frac{21}{4}\ \dots(4)$Subtracting (3) from (4), we get
$-\frac{1}{4}\text{x}=-\frac{3}4{}$ $-\text{x}=-3$ $\Rightarrow\text{x}=3$Substitution x = 3 in (1), we get
$2\times3-\frac{\text{3y}}{4}=3$ $-\frac{\text{3y}}{4}=3-6$ $-\frac{\text{3y}}{4}=-3$ $\Rightarrow\text{y}=\frac{-3\times4}{-3}=4$$\therefore$ solution is x = 3 and y = 4
View full question & answer→