Question

3a + 4b = 43                      -2a + 3b = 11 (BY ELIMINATION)​

Answer 1

$\large{\underline{\sf{\orange{Given-}}}}$

• $\sf{3a + 4b = 43 \: .... \: (1)}$

• $\sf{ - 2a + 3b = 11 \: .... \: (2)}$

$\large{\underline{\sf{\purple{To \: Find-}}}}$

• Values of a & b

$\large{\underline{\sf{\blue{Solution-}}}}$

Multiplying the Equation (1) by (×3) & Multiplying the Equation (2) by (×4)

$\sf\leadsto{3a + 4b = 43 \: .... \: ( \times 3)}$

$\sf\leadsto{ - 2a + 3b = 11 \: .... \: ( \times 4)}$

We get,

$\sf\leadsto{9a \: {\cancel{+12b}} = 129}$

$\sf\leadsto{ - 8a \: {\cancel{+12b}} = 44}$

$\: \: \: \: \: \: \: {( + ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( - )}$

______________________

$\sf\leadsto{17a = 85}$

$\sf\leadsto{a = \frac{\cancel{85}}{\cancel{17}}}$

$\boxed{\sf\red{★ \: a = 5}}$

Placing (a = 5) in the Equation (2),

$\sf\leadsto{ - 2a + 3b = 11 \: .... \: (2)}$

$\sf\leadsto{ - 2 \times 5 + 3b = 11}$

$\sf\leadsto{ - 10 + 3b = 11}$

$\sf\leadsto{3b = 11 + 10}$

$\sf\leadsto{3b = 21}$

$\sf\leadsto{b = \frac{\cancel{21}}{\cancel{3}}}$

$\boxed{\sf\green{★ \: b = 7}}$

Hence,

• a = 5

• b = 7