Q7. - Algebra 1

Question

Use elimination method to solve the system of equations below:

$\begin{array}{c}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{2}\\ \mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{18}\end{array}$

A $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}$               B $\mathbf{(}\mathbf{7}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{)}$                 C $\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}$               D $\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{5}\mathbf{)}$ 

Step-by-Step Solution

Answer

The solution of the given system of equations is option D $\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{5}\mathbf{)}$.

Step 1. State the concept of Linear System with two variables.

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.

Step 2. State the concept of ‘Elimination Method’.

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.

Step 3. Solve the given system of equations.

The given system of equations are,

$\begin{array}{c}3x+2y=-2\dots \left(1\right)\\ 2x-2y=-18\dots \left(2\right)\end{array}$

Notice that the coefficient of variable $‘y’$ is 2 in both the equation (1) and (2).

Since in both the equations the coefficient of $y$ are same and sign of variable $‘y’$ are different , eliminate variable $‘y’$ by adding equations (1) and (2).

$\begin{array}{c}(3x+2y)+(2x-2y)=-2+(-18)\\ 3x+2y+2x-2y=-2-18\\ 3x+2x+2y-2y=-20\\ 5x=-20\\ \frac{5x}{5}=\frac{-20}{5}\\ x=-4\end{array}$

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

 

Substituting $x=-4$ in equation (1)

 $\begin{array}{c}3(-4)+2y=-2\\ -12+2y=-2\\ 2y=-2+12\\ 2y=10\\ \frac{2y}{2}=\frac{10}{2}\\ y=5\end{array}$

Therefore, the solution is $x=-4$ and $y=5$.

 

Hence option D $(-4,5)$ is correct.