91Ó°ÊÓ

A linear system of two equations with two variables is any system that can be written in the form.

$\begin{array}{l}ax+by=p\\ cx+dy=q\end{array}$

Where any of the constants can be zero with the exception that each equation must have at least one variable in it.

Also, the system is called linear if the variables are only to the first power, are only in the numerator and there are no products of variables in any of the equations.

In the elimination method you either add or subtract the equations to get an equation in one variable.

When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

The given system of equations are,

$\begin{array}{l}3x−8y=−50\text{ } \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}…\left(1\right)\\ 3x−5y=−38\text{ } \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰}…\left(2\right)\end{array}$

Note that the coefficient of variable $‘x’$is 3 in both the equation (1) and (2).

Since in both the equations the coefficient and sign of variable $‘x’$are same, eliminate variable $‘x’$by subtracting equation (2) from (1).

$\begin{array}{c}(3x−8y)−(3x−5y)=−50−(−38)\\ 3x−8y−3x+5y=−50+38\\ 3x−3x−8y+5y=−12\\ −3y=−12\\ \frac{−3y}{−3}=\frac{−12}{−3}\\ y=4\end{array}$

Substitute $y=4$, in either of the original equations to get the value of $‘x’$.

Substitute $y=4$in $3x−8y=−50$ to find the value of $\text{'}x\text{'}$.

$\begin{array}{c}3x−8y=−50\\ 3x−8\left(4\right)=−50\\ 3x−32=−50\\ 3x=−50+32\\ 3x=−18\\ \frac{3x}{3}=−\frac{18}{3}\\ x=−6\end{array}$

Therefore, the solution of the given system of equations is $x=−6$and $y=4$.

Hence, option D $(−6,4)$is correct.