NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions Ex 3.4 Math Solution

By elimination method
\(3 x+4 y=10\ldots(i)\)
\(2 x-2 y=2\ldots(ii)\)
Multiplying equation (ii) by 2 we get,
\(
4 x-4 y=4\ldots(iii)\)
Adding equations (i) and (iii) we get,
\(7 x=14\)
\(=x=\frac{14}{7}=2\)
Putting value of \( x \) in equation (i) we get,
\(3(2)+4 y=10\)
\(4 y=10-6=4\)
\(y=\frac{4}{4}=1\)
By substitution method
\(3 x+4 y=10\ldots(i)\)
\(2 x-2 y=2\ldots(ii)\)
From equation (ii)
\(
2 x=2+2 y
\)
Dividing both side by 2 , we getx \( =1+\mathrm{y} \)
putting value of \( x \) in equation (i) we get,
\(3(1+y)+4 y=10\)
\(3+3 y+4 y=107 y=7 y=1\)
and
\( \mathrm{x}=1+\mathrm{y}=1+1=2 \)

By elimination method
\(
3 x-5 y=4\ldots(i)\)
\(
9 x-2 y=7\ldots(ii)\)
Multiplying equation (i) by 3 we get,
\(
9 x-15 y=12\ldots(iii)\)
Subtracting equation (ii) from (iii) we get,
\(-13 y=5\)
\(=y=-\frac{5}{13}\)
Putting value of \( y \) in equation (i) we get,
\(3 x-5 \times\left(-\frac{5}{3}\right)=4\)
\(3 x+\frac{25}{13}=4\)
Taking LCM,
\(
3 x=\frac{4}{1}-\frac{25}{13}=\frac{52-25}{13}=\frac{27}{13}\)
\(\text { or } x=\frac{27}{13 \times 3}=\frac{9}{13}\)
By substitution method
\(3 x-5 y=4\ldots(i)\)
\(9 x-2 y=7\ldots(ii)\)
From equation (i)
\(
x=\frac{4+5 y}{3}
\)
Putting value of \(x\) in equation (ii) we get,
\(=9 \times \frac{4+5 y}{3}-2 y=7\)
\(=12+15 y-2 y=7\)
\(=13 y=-5\)
\(y=-\frac{5}{13}\)
Putting value of \( y \) in equation (i) we get
\(3 x-5 \times-\frac{5}{13}=4\)
\(3 x=4-\frac{25}{13}=\frac{27}{13}\)
\(=x=\frac{27}{39}=\frac{9}{13}\)