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Maths 2 Marks Questions

Linear Equations In Two Variables · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 6 questions.

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Question 12 Marks
Which value(s) of $\lambda,$ do the pair of linear equations $\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$have:
Infinitely many solutions?
Answer
For infinitely many solutions,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\lambda}{1}=\frac{1}{\lambda}=\frac{\lambda^2}{1}$
$\Rightarrow\frac{\lambda}{1}=\frac{\lambda^2}{1}$
$\Rightarrow\lambda(\lambda-1)=0$
When $\lambda\neq0,$ then $\lambda=1$
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Question 22 Marks
On comparing the ratios $\frac{\text{a}_1}{\text{a}_2},\frac{\text{b}_1}{\text{b}_2}$ and $\frac{\text{c}_1}{\text{c}_2},$ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.
$5x - 4y + 8 = 0$
$7x + 6y - 9 = 0$
Answer
Given equation are:
$5x - 4y + 8 = 0$
$7x + 6y - 9 = 0$
Where,$a_1 = 5, b_1 = -4, c_1= 8$
$a_2 = 7, b_2 = 6, c_3= -9$
We get $\frac{\text{a}_1}{\text{a}_2}=\frac{5}{7},\frac{\text{b}_1}{\text{b}_2}=\frac{-4}{6}=\frac{-2}{3}$
and $\frac{\text{c}_1}{\text{c}_2}=\frac{8}{-9}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Thus the pair of linear equation is intersecting.
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Question 32 Marks
Which value(s) of $\lambda,$ do the pair of linear equations $\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$have:
a unique solutions?
Answer
For a Unique solution,$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\Rightarrow\frac{\lambda}{1}\neq\frac{1}{\lambda}$
$\Rightarrow\lambda^2\neq1$
$\Rightarrow\lambda\neq\pm1$
So, all real values of $\lambda$ except $\pm1$
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Question 42 Marks
On comparing the ratios $\frac{\text{a}_1}{\text{a}_2},\frac{\text{b}_1}{\text{b}_2}$ and $\frac{\text{c}_1}{\text{c}_2},$ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.
$6x - 3y + 10 = 0$
$2x - y + 9 = 0$
Answer
Given equation are:
$6x - 3y + 10 = 0$
$2x - y + 9 = 0$
Where, $a_1 = 6, b_1 = -3, c_1= 10$
$a_2 = 2, b_2 = -1, c_3= 9$
We get $\frac{\text{a}_1}{\text{a}_2}=\frac{6}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{-1}=\frac{\text{c}_1}{\text{c}_2}=\frac{10}{9}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=3$
Thus the pair of line is parallel lines.
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Question 52 Marks
On comparing the ratios $\frac{\text{a}_1}{\text{a}_2},\frac{\text{b}_1}{\text{b}_2}$ and $\frac{\text{c}_1}{\text{c}_2},$ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.
$9x + 3y + 12 = 0$
$18x + 6y + 24 = 0$
Answer
Given equation are:
$9x + 3y + 12 = 0$
$18x + 6y + 24 = 0$
Where, $a_1 = 9, b_1 = 3, c_1= 12$
$a_2 = 18, b_2 = 6, c_3= 24$
We get $\frac{\text{a}_1}{\text{a}_2}=\frac{9}{18},\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}$
and $\frac{\text{c}_1}{\text{c}_2}=\frac{12}{24}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Thus the pair of linear equation are coincide.
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Question 62 Marks
Solve the following systems of equations:
$\frac{4}{\text{x}}+3\text{y}=8,$
$\frac{6}{\text{x}}-4\text{y}=-5.$
Answer
$\frac{4}{\text{x}}+3\text{y}=8\ ......(\text{i})$
$\frac{6}{\text{x}}-4\text{y}=-5\ ......(\text{ii})$
equations (i) × 3
$\Rightarrow\frac{12}{\text{x}} + 9\text{y} = 24\ ...(\text{iii})$
and euations (ii) × 2
$\Rightarrow\frac{12}{\text{x}} - 8\text{y} - 10\ ...(\text{iv})$
Substracting (iv) frpm (iii)
$\frac{12}{\text{x}}+9\text{y}-\frac{12}{\text{x}}+8\text{y}=24+10$
$17\text{y}=34$
$\text{y}=2$
Putting y in (iii) equation
$\frac{12}{\text{x}}+9\times2=24$
$\frac{12}{\text{x}}=24-18$
$\frac{12}{\text{x}}=6$
$\text{x}=2$
So that solution is x = 2, y = 2.
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