Question

Solve the following system of equations using elimination method.

$3x+4y=24,20x-11y=47$

• A. $(4,3)$
• B. $(2,3)$
• C. $(4,2)$
• D. None of these

Answer 1

Answer: (A)

$(4,3)$

Consider the given equations.

$3x+4y=24$ $-------\left(1\right)$

$20x-11y=47$ $-------\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

On multiplying by $20$ in equation $\left(1\right)$ and multiplying by $3$ in equation $\left(2\right)$ and subtract, we get

$60x+80y=480$

$60x-33y=141$

—————————

$113y=339$

$y=3$ $[$ put in $\left(2\right)]$

$20x-11\times 3=47$

$20x-33=47$

$20x=80$

$x=4$

Hence, the value of $x$ is $4$ and the value of $y$ is $3$.