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 $a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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 $a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$\text { Where } a_1=2, b_1=-3, c_1=-5$
$a_2=6, b_2=-9, c_2=-15$
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{6}=\frac{1}{3},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-3}{-9}=\frac{1}{3}$ 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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$\text { Where } a_1=k, b_1=3, c_1=-(k-3)$
$a_2=12, b_2=k, c_2=-k$
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 $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$
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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$\text { Where } a_1=5, b_1=2, c_1=-2 k$
$a_2=2(k+1), b_2=k, c_2=-(3 k+4)$
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:
$\frac{\text{k}-1}{\text{k}+1}=\frac{1}{(\text{k}-1)}$
$ (\mathrm{k}-1)^2=\mathrm{k}+1 $
$ \Rightarrow \mathrm{k}^2+1-2 \mathrm{k}=\mathrm{k}+1 $
$ \Rightarrow \mathrm{k}^2+1-1-2 \mathrm{k}-\mathrm{k}=0 $
$ \Rightarrow \mathrm{k}^2=3 \mathrm{k} $
$ \Rightarrow \mathrm{k}=3$
Case II:
$\frac{1}{(\text{k}-1)}=\frac{5}{(\text{3k}+1)}$
$(3k + 1) = 5(k - 1)$
$⇒ 3k + 1 = 5k - 5$
$⇒ -2k = -6$
$⇒ k = 3$
Case III:
$\frac{\text{k}-1}{\text{k}+1}=\frac{5}{(\text{3k}+1)}$
$ \Rightarrow(\mathrm{k}-1)(3 \mathrm{k}+1)=5(\mathrm{k}+1) $
$ \Rightarrow 3 \mathrm{k}^2+\mathrm{k}-3 \mathrm{k}-1=5 \mathrm{k}+5 $
$ \Rightarrow 3 \mathrm{k}^2-2 \mathrm{k}-5 \mathrm{k}-1-5=0 $
$ \Rightarrow 3 \mathrm{k}^2-7 \mathrm{k}-6=0 $
$ \Rightarrow 3 \mathrm{k}^2-9 \mathrm{k}+2 \mathrm{k}-6=0 $
$ \Rightarrow 3 \mathrm{k}(\mathrm{k}-3)+2(\mathrm{k}-3)=0 $
$ \Rightarrow(3 \mathrm{k}+2)(\mathrm{k}-3)=0 $
$ \Rightarrow 3 \mathrm{k}=-2 \text { or } \mathrm{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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$\text { Where } a_1=(a-1), b_1=3, c_1=-2$
$a_2=6, b_2=(1-2 b), c_2=-6$
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 $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$, 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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$.
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
$⇒ a + b = 3a$
$⇒ 4 + b = 12$
$⇒ b = 8$
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$kx + 3y = k - 3, 12x + ky = k$
$⇒ kx + 3y - (k - 3) = 0, 12x + ky - k = 0$
We know that,
the system of linear equations
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0 $
$ \text { Where } a_1=(2 a-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 $a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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 $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$,
where $a_1=2, b_1=3, c_1=-7, a_2=k-1, b_2=k+2$ and $c_2=-3 k$
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 $=\frac{2}{\text{k}-1}=\frac{-7}{-\text{3k}}$
$⇒ 2(k + 2) = 3(k - 1), 9k = 7k + 14$ and $6k = 7k - 7$
$⇒ k = 7, k = 7$ 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_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0 $
$\text { Where } a_1=5, b_1=2, c_1=-2 k $
$a_2=2(k+1), b_2=k, c_2=-(3 k+4)$
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
$\frac{\text{5}}{2(\text{k}+1)}=\frac{2}{\text{k}}$
$⇒ 5k = 4(k + 1)$
$⇒ 5k = 4k + 4$
$⇒ k = 4$
Case: II
$\frac{\text{2}}{\text{k}}=\frac{\text{2k}}{(\text{3k}+4)}$
$\Rightarrow 2(3 \mathrm{k}+4)=2 \mathrm{k}^2$
$\Rightarrow 6 \mathrm{k}+8=2 \mathrm{k}^2$
$\Rightarrow 2 \mathrm{k}^2-6 \mathrm{k}-8=0$
$\Rightarrow 2\left(\mathrm{k}^2-3 \mathrm{k}-4\right)=0$
$\Rightarrow \mathrm{k}^2-3 \mathrm{k}-4=0$
$\Rightarrow \mathrm{k}^2-4 \mathrm{k}+\mathrm{k}-4=0$
$\Rightarrow \mathrm{k}(\mathrm{k}-4)+1(\mathrm{k}-4)=0$
$\Rightarrow(\mathrm{k}-4)(\mathrm{k}+1)=0$
$\Rightarrow(\mathrm{k}-4)=0 \text { or } \mathrm{k}+1=0$
$\Rightarrow \mathrm{k}=4 \text { or } \mathrm{k}=-1$
Case: III
$\frac{\text{5}}{2(\text{k}+1)}=\frac{\text{2k}}{(\text{3k}+4)}$
$ \Rightarrow 15 \mathrm{k}+20=4 \mathrm{k}^2=4 \mathrm{k}$
$\Rightarrow 4 \mathrm{k}^2+4 \mathrm{k}-15 \mathrm{k}-20=0$
$\Rightarrow 4 \mathrm{k}^2-11 \mathrm{k}-20=0$
$\Rightarrow 4 \mathrm{k}^2-16 \mathrm{k}+5 \mathrm{k}-20=0$
$\Rightarrow 4 \mathrm{k}(\mathrm{k}-4)+5(\mathrm{k}-4)=0$
$\Rightarrow(\mathrm{k}-4)(4 \mathrm{k}+5)=0$
$⇒ k = 4$ 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
$7x = 42$
$⇒ x = 6$
Putting $x = 6$ in $(1),$ we get
$2 × 6 + 3y = 18$
$⇒ 3y = 18 - 12$
$⇒ 3y = 6$
$\Rightarrow\text{y}=\frac{6}{3}=2$
$\therefore$ 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_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0 $
$ \text { Where } a_1=k, b_1=3, c_1=-(2 k+1) $
$ a_2=2(k+1), b_2=9, c_2=-(7 k+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
$\frac{\text{k}}{2(\text{k}+1)}=\frac{1}{\text{3}}$
$⇒ 2(k + 1) = 3k$
$⇒ 2k + 2 = 3k$
$⇒ k = 2$
Case: II
$\frac{\text{1}}{\text{3}}=\frac{\text{2k}+1}{\text{7k}+1}$
$⇒ 7k + 1 = 6k + 3$
$⇒ 7k - 6k = 3 - 1$
$⇒ k = 2$
Case: III
$\frac{\text{k}}{2(\text{k}+1)}=\frac{\text{2k}+1}{\text{7k}+1}$
$ \Rightarrow k(7 k+1)=2(2 k+1)(k+1) $
$ \Rightarrow 7 k^2+k=2\left(2 k^2+2 k+k+1\right) $
$ \Rightarrow 7 k^2+k=2\left(2 k^2+3 k+1\right) $
$ \Rightarrow 7 k^2+k=4 k^2+6 k+2 $
$ \Rightarrow 7 k^2-4 k^2+k-6 k-2=0 $
$ \Rightarrow 3 k^2-5 k-2=0 $
$ \Rightarrow 3 k^2-(6 k-1 k)-2=0 $
$ \Rightarrow 3 k(k-2)+1(k-2)=0 $
$ \Rightarrow(k-2)(3 k+1)=0$
$⇒ k = 2$ 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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$\text { Where } a_1=2, b_1=3, c_1=-7$
$a_2=(k-1), b_2=(k+2), c_2=-3 k$
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
$\frac{2}{\text{k}-1}=\frac{3}{\text{k}+2}$
$⇒ 2(k + 2) = 3(k - 1)$
$⇒ 2k + 4 = 3k - 3$
$⇒ k = 7$
Case: II
$\frac{3}{\text{k}+2}=\frac{7}{\text{3k}}$
$⇒ 7(k + 2) = 9k$
$⇒ 7k + 14 = 9k$
$⇒ k = 7$
Case: III
$\frac{2}{\text{k}-1}=\frac{7}{\text{3k}}$
$⇒ 7k - 7 = 6k$
$⇒ k = 7$
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 $y:$
$2x - y + 3 = 0,$
$3x - 7y + 10 = 0$
Answer$2x - y + 3 = 0$
$⇒ 2x - y = - 3 ...(i)$
$3x - 7y + 10 = 0$
$⇒ 3x - 7y = 10 ...(ii)$
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 $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$,
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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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 $7(a + 2b + 2) = 3(4(a + b) + 1)$
$⇒ 2a + 4b + 4 = 3a + 3b + 3$ and $7a + 14b + 14 = 12a + 12b + 3$
$⇒ a - b = 1 ...(i)$ and $5a - 2b = 11 ...(ii)$
Multiply $(i)$ by $2$ and subtract from $(ii).$
$⇒ 2a - 2b = 2$ and $5a - 2b = 11$
$3a = 9$
$⇒ a = 3$
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 $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$,
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 $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$,
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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$\text { Where } a_1=2, b_1=(k-2), c_1=-k$
$a_2=6, b_2=(2 k-1), c_2=(2 k+5)$
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
$\frac{1}{\text{3}}=\frac{\text{k}-2}{2\text{k}-1}$
$⇒ 2k - 1 = 3k - 6$
$⇒ k = 5$
Case: II
$\frac{\text{k}-2}{\text{2k}-1}=\frac{\text{k}}{\text{2k}+5}$
$⇒ (k - 2)(2k + 5) = k(2k - 1)$
$⇒ k + k = 10$
$⇒ 2k = 10$
$\Rightarrow\text{k}=\frac{10}{2}=5$
Case: III
$\frac{1}{\text{3}}=\frac{\text{k}}{\text{2k}+5}$
$⇒ 2k + 5 = 3k$
$⇒ 3k - 2k = 5$
$⇒ k = 5$
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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$\text { Where } a_1=2, b_1=-3, c_1=-7$
$a_2=(a+b), b_2=(a+b-3), c_2=-(4 a+b)$
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$kx + 3y = 3, 12x + ky = 6$
$⇒ kx + 3y - 3 = 0, 12x + ky - 6 = 0$
We know that,
the system of linear equations
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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_1 x+b_1 y+c_1=0, a_2 x+b_2 y+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_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0 $
$\text { Where } a_1=5, b_1=2, c_1=-2 k$
$a_2=2(k+1), b_2=k, c_2=-(3 k+4)$
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
$\frac{\text{5}}{2(\text{k}+1)}=\frac{2}{\text{k}}$
$⇒ 5k = 4(k + 1)$
$⇒ 5k = 4k + 4$
$⇒ k = 4$
Case: II
$\frac{\text{2}}{\text{k}}=\frac{\text{2k}}{(\text{3k}+4)}$
$\Rightarrow 2(3 \mathrm{k}+4)=2 \mathrm{k}^2$
$\Rightarrow 6 \mathrm{k}+8=2 \mathrm{k}^2$
$\Rightarrow 2 \mathrm{k}^2-6 \mathrm{k}-8=0$
$\Rightarrow 2\left(\mathrm{k}^2-3 \mathrm{k}-4\right)=0$
$\Rightarrow \mathrm{k}^2-3 \mathrm{k}-4=0$
$\Rightarrow \mathrm{k}^2-4 \mathrm{k}+\mathrm{k}-4=0$
$\Rightarrow \mathrm{k}(\mathrm{k}-4)+1(\mathrm{k}-4)=0$
$\Rightarrow(\mathrm{k}-4)(\mathrm{k}+1)=0$
$\Rightarrow(\mathrm{k}-4)=0 \text { or } \mathrm{k}+1=0$
$\Rightarrow \mathrm{k}=4 \text { or } \mathrm{k}=-1$
Case: III
$\frac{\text{5}}{2(\text{k}+1)}=\frac{\text{2k}}{(\text{3k}+4)}$
$ \Rightarrow 15 \mathrm{k}+20=4 \mathrm{k}^2=4 \mathrm{k}$
$ \Rightarrow 4 \mathrm{k}^2+4 \mathrm{k}-15 \mathrm{k}-20=0$
$ \Rightarrow 4 \mathrm{k}^2-11 \mathrm{k}-20=0$
$ \Rightarrow 4 \mathrm{k}^2-16 \mathrm{k}+5 \mathrm{k}-20=0$
$ \Rightarrow 4 \mathrm{k}(\mathrm{k}-4)+5(\mathrm{k}-4)=0$
$ \Rightarrow(\mathrm{k}-4)(4 \mathrm{k}+5)=0$
$⇒ k = 4$ 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 $a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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$8x + 5y = 9, kx + 10y = 15$
$⇒ 8x + 5y - 9 = 0, kx + 10y - 15 = 0$
We know that,
the system of linear equations
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0 $
$\text { 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_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$ \text { 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?
Answer
- $kx + 2y - 5 = 0,$
$3x - 4y - 10 = 0$
These equations are of the form
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
$ \text { 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.
- 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 $a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$ 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.
$⇒ (k - 3)x + 3y - k = 0, kx + ky - 12 = 0$
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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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
$a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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 $a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
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→