Question

Solve the given pair of equations by substitution method:

$2a+3b=6$
$3a+5b=15$

• A. $a=3,b=2$
• B. $a=-15,b=12$
• C. $a=8,b=15$
• D. $a=-7,b=3$

Answer 1

Answer: (B)

$a=-15,b=12$

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $2a+3b=6$ and write it as

$b=\frac{6-2a}{3}.........\left(1\right)$

Substitute the value of $b$ in Equation $3a+5b=15$. We get

$3a+5\left(\frac{6-2a}{3}\right)=15\Rightarrow 9a+5(6-2a)=45\Rightarrow 9a+30-10a=45\Rightarrow -a=45-30\Rightarrow a=-15$

Therefore, $a=-15$

Substituting this value of $a$ in Equation (1), we get

$b=\frac{6-(-30)}{3}$

$b=\frac{36}{3}=12$

Hence, the solution is $a=-15,b=12$.