Q20. - Algebra 2

Question

Find the point at which the graphs of $\mathbf{2}x+\mathbf{3}y=\mathbf{7}$and $\mathbf{3}x-\mathbf{4}y=\mathbf{2}$ intersect.

Step-by-Step Solution

Answer

The graph intersects at the point $\left(2,1\right).$

Step 1 – Use elimination method

The given equations are,

               $\begin{array}{l}2x+3y=7\\ 3x-4y=2\end{array}$

              $\begin{array}{r}\mathrm{....}\left(1\right)\\ \mathrm{....}\left(2\right)\end{array}$

Multiply (1) by 3 and multiply (2) by 2, we get

$\begin{array}{c}6x+9y=21\\ 6x-8y=4\end{array}$

It can be rewritten as follows,

$\begin{array}{c}6x+9y-21=0\\ 6x-8y-4=0\end{array}$

$\begin{array}{r}\mathrm{....}\left(3\right)\\ \mathrm{....}\left(4\right)\end{array}$

Subtracting equation (4) from (3), we get

$\begin{array}{c}(6x+9y-21)-(6x-8y-4)=0\\ 6x+9y-(6x-8y)=21-4\\ 6x+9y-6x+8y=17\\ 17y=17\\ y=\frac{17}{17}\\ y=1\end{array}$

Step 2 – Substitute the value of y in equation (1)

We know that $y=1.$

Substituting the value, we get

$\begin{array}{c}2x+3y=7\\ 2x+3\left(1\right)=7\\ 2x+3=7\\ 2x=7-3\\ 2x=4\\ x=2\end{array}$

Thus, the point at which the graph intersects is $(2,1).$