Question

Solve the given pair of equations by substitution method:

$4a-b=10$
$2a+3b=12$

• A. $a=0,b=5$
• B. $a=7,b=6$
• C. $a=1,b=7$
• D. $a=3,b=2$

Answer 1

The correct option is D $a=3,b=2$

We have, $4a-b=10\Rightarrow b=4a-10$.........(1)

and $2a+3b=12$........(2)

Now substitute value of b in terms of a in (2)

$\Rightarrow 2a+3(4a-10)=12$

$\Rightarrow 2a+12a-30=12$

$\Rightarrow 14a=12+30=42\Rightarrow a=\frac{42}{14}=3$

So $b=4a-10=4\left(3\right)-10=12-10=2$

Hence option 'D' is correct choice