Question

Which of the following is the solution set for the simultaneous equation ?
$\frac{2x-1}{3}=\frac{y+6}{5}=\frac{3x-y}{7}$

• A. $x=2;y=-1$
• B. $x=3;y=-1$
• C. $x=-1;y=2$
• D. $x=4;y=1$

Answer 1

Answer: (A)

$x=2;y=-1$

Given : $\frac{2x-1}{3}=\frac{y+6}{5}=\frac{3x-y}{7}$

Take two equalities simultaneously we get

$\frac{2x-1}{3}=\frac{y+6}{5}$ and $\frac{y+6}{5}=\frac{3x-y}{7}$

$⟹$ $10x-5=3y+18$ and $7y+42=15x-5y$

$⟹$ $10x-3y=23$ .......... (1) and $15x-12y=42$ .......... (2)

from eq(2) we can get the value of $x$ in terms of $y$ i.e, $x=\frac{42+12y}{15}$

now simply substitute this $x$ value in eq(1) from where we can calculate the value of $y$ and then substitute the $y$ in any eq we can find the value of $x$ , so

Solving (1) and (2) simultaneously we get

$x=2$

Substituting $x$ in (1) we get $y=-1$.

Hence, $x=2;y=-1$.