Solve graphically the system of equations 4x – y – 4 = 0 3x + 2y - 14 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Home » Solve graphically the system of equations 4x – y – 4 = 0 3x + 2y – 14 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solve graphically the system of equations
4x – y – 4 = 0
3x + 2y – 14 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solve graphically the system of equations
4x – y – 4 = 0
3x + 2y – 14 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solution:

Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line $Y O Y^{\prime}$ as the $x$ -axis and $y$ -axis, respectively.
Graph of $4 x-y=4$
$\begin{array}{l} 4 x-y=4 \\ \Rightarrow y=(4 x-4) \ldots \ldots . . \text { (i) } \end{array}$

Putting $x=0$ , we obtain $y=-4$ .
Putting $x=1$ , we obtain $y=0$ .
Putting $x=2$ , we obtain $y=4$

Therefore, we have the following table for the equation $4 x-y=4$ .

$\begin{tabular}{|c|c|c|c|} \hline$x$ & 0 & 1 & 2 \\ \hline$y$ & $-4$ & 0 & 4 \\ \hline \end{tabular}$

Now, plot the points $A(0,-4), B(1,0)$ and $C(2,4)$ on the graph paper. Join $A B$ and $B C$ to get the graph line $A C$ . Extend it on both ways.

Therefore, $A C$ is the graph of $4 x-y=4$
Graph of $3 x+2 y=14$
$\begin{array}{l} 3 x+2 y=14 \\ \Rightarrow 2 y=(14-3 x) \\ \therefore y=\frac{14-3 x}{2} \quad \ldots \ldots \ldots . \end{array}$

Putting $x=0$ , we obtain $y=7$ .
Putting $x=2$ , we obtain $y=4$ .
Putting $x=4$ we obtain $y=1$ .

So, we have the following table for the equation $3 x+2 y=14$ .

$\begin{tabular}{|c|c|c|c|} \hline$x$ & 0 & 2 & 4 \\ \hline$y$ & 7 & 4 & 1 \\ \hline \end{tabular}$

Now, plots the points $P(0,7)$ and $Q(4,1)$ . The point $C(2,4)$ has already been plotted. Join $P C$ and $C Q$ to get the graph line $P Q$ .
Extend it on both ways.