Question

Solve the following system of equations graphically. Also, find the point where the lines meet the $X-$ axis.
$\begin{align}
  & x+2y=5 \\
 & 2x-3y=-4 \\
\end{align}$

Answer 1

- Hint: For solving this question, first we will see the general equation of the straight line $y=mx+c$ of slope $m$ and it intersects $Y-$ the axis at the point $(0,c)$ and it intersects $X-$ axis at the point $\left( -\dfrac{c}{m},0 \right)$ . After that, we will write the given equations in the form of $y=mx+c$ and we will try to find at least two points on each line. Then, we will plot them on the graph and find their point of intersection and the points where the lines meet the $X-$ axis.


Complete step-by-step solution -

Given:
We have the following equations:
$\begin{align}
  & x+2y=5 \\
 & 2x-3y=-4 \\
\end{align}$
Now, it is given that we have to solve the given equations graphically and also find where the lines meet the $X-$ axis.
Now, as we know that, any equation of the form $y=mx+c$ is the equation of a straight line of slope $m$ and it intersects $Y-$ axis at the point $(0,c)$ and it intersects $X-$ axis at the point $\left( -\dfrac{c}{m},0 \right)$ .
Now, let ${{L}_{1}}:x+2y=5$ and ${{L}_{2}}=2x-3y=-4$ .
Now, we will write the equation of lines ${{L}_{1}}$ and ${{L}_{2}}$ into the form of $y=mx+c$ and then find the value of $m$ and $c$ for them. Then,
$\begin{align}
  & {{L}_{1}}:x+2y=5 \\
 & \Rightarrow {{L}_{1}}:2y=-x+5 \\
 & \Rightarrow {{L}_{1}}:y=-\dfrac{1}{2}x+\dfrac{5}{2}................\left( 1 \right) \\
 & {{L}_{2}}=2x-3y=-4 \\
 & \Rightarrow {{L}_{2}}=-3y=-2x-4 \\
 & \Rightarrow {{L}_{2}}=y=\dfrac{2}{3}x+\dfrac{4}{3}.................\left( 2 \right) \\
\end{align}$
Now, we will compare equation (1) and (2) with the general equation of the straight line $y=mx+c$ . Then,
$\begin{align}
  & {{L}_{1}}:y=-\dfrac{1}{2}x+\dfrac{5}{2}={{m}_{1}}x+{{c}_{1}} \\
 & \Rightarrow {{m}_{1}}=-\dfrac{1}{2}\text{ and }{{c}_{1}}=\dfrac{5}{2} \\
 & {{L}_{2}}=y=\dfrac{2}{3}x+\dfrac{4}{3}={{m}_{2}}x+{{c}_{2}} \\
 & \Rightarrow {{m}_{2}}=\dfrac{2}{3}\text{ and }{{c}_{2}}=\dfrac{4}{3} \\
\end{align}$
Now, from the above result, we conclude that line ${{L}_{1}}:y=-\dfrac{1}{2}x+\dfrac{5}{2}$ passes through points $A= \left( 0,{{c}_{1}} \right)=\left( 0,\dfrac{5}{2} \right)\And B=\left( -\dfrac{{{c}_{1}}}{{{m}_{1}}},0 \right)= \left( 5,0 \right)$ and, the line ${{L}_{2}}=y=\dfrac{2}{3}x+\dfrac{4}{3}$ passes through points $C= \left( 0,{{c}_{2}} \right)= \left( 0,\dfrac{4}{3} \right)\And D= \left( -\dfrac{{{c}_{2}}}{{{m}_{2}}},0 \right)=\left( -2,0 \right)$ .
Now, we can plot the given lines on the graph easily. For more clarity, look at the figure given below:

Now, from the above figure, it is evident that both lines intersect at a point $E= \left( 1,2 \right)$ . Moreover, the line ${{L}_{1}}:x+2y=5$ intersects $X-$ axis at point $B= \left( 5,0 \right)$ and line ${{L}_{2}}=2x-3y=-4$ intersect $X-$ axis at point $D= \left( -2,0 \right)$ .
Now, we can check whether point $E= \left( 1,2 \right)$ satisfies the equation of ${{L}_{1}}:x+2y=5$ and ${{L}_{2}}=2x-3y=-4$ . Then,
$\begin{align}
  & {{L}_{1}}:x+2y=5 \\
 & \Rightarrow 1+2\times 2=5 \\
 & \Rightarrow 1+4=5 \\
 & \Rightarrow 5=5 \\
 & {{L}_{2}}=2x-3y=-4 \\
 & \Rightarrow 2\times 1-3\times 2=-4 \\
 & \Rightarrow 2-6=-4 \\
 & \Rightarrow -4=4 \\
\end{align}$
Now, from the above result, it is verified that both lines intersect at a point $E= \left( 1,2 \right)$ .
Thus, if $x+2y=5$ and $2x-3y=-4$ , then $x=1$ and $y=2$ .

Note: Here, the student should apply the concepts of coordinate geometry correctly and though the problem is very easy, we should avoid calculation mistakes while solving. Moreover, in such questions, we can also solve the equations directly and then, we should verify it from graphs for better understanding.