Question

Find the coordinates of the points of intersection of the straight lines whose equations are: $2x-3y+5=0$ and $7x+4y=3$ .

Answer 1

- Hint: The given equations in the question are linear equations. We can find the coordinates of the points of intersection of the straight lines by finding the values of x and y using elimination method. First write both the equations in the form of $ax+by=c$ . Then we make the coefficients of one of the variables, say ‘x’, equal by multiplying each equation by suitable real numbers. Then, we can perform addition or subtraction of two equations to eliminate one variable. We can solve the equation for the remaining variable. Substituting the obtained variable in any one equation, we can find the other.

Complete step-by-step solution -

To find the coordinates of the points of intersection of the straight lines for the given equations are: $2x-3y+5=0$ and $7x+4y=3$ .
First we will make the $2x-3y+5=0$ equation in the form of $ax+by=c$ .
Subtracting ‘5’ on both sides, we get –
$2x-3y=-5$
Let us consider $2x-3y=-5$ as equation (1) and $7x+4y=3$ as equation (2)
Let us eliminate ‘y’ from the given equations. The coefficients of ‘y’ in the given equations are -3 and 4.
So, we will multiply equation (1) by 4 and equation (2) by -3.
Equation $\left( 1 \right)\times 4=\left( 2x-3y \right)\times 4=-5\times 4$
$8x-12y=-20$ …………………………….. (3)
Equation $\left( 2 \right)\times -3=\left( 7x+4y \right)\times -3=3\times -3$
$-21x-12y=-9$ ……………………………….. (4)
Now, we will subtract equation (3) from (4) as to eliminate ‘y’ and find the value of ‘x’.
$\begin{align}
  & -21x-12y=-9 \\
 & \text{ }8x\text{ }-12y=-20 \\
 & \dfrac{\left( - \right)\text{ }\left( + \right)\text{ }\left( + \right)}{-29x\text{ =11}} \\
\end{align}$
Therefore, we get $-29x=11$ .
By dividing ‘29’ on both sides, we get –
$-x=\dfrac{11}{29}\text{ or }x=-\dfrac{11}{29}$
Thus, we can substitute the value of x in equation (1) to get y coordinates –
$\begin{align}
  & 2x-3y=-5 \\
 & 2\left( -\dfrac{11}{29} \right)-3y=-5 \\
 & -\dfrac{22}{29}-3y=-5 \\
\end{align}$
Adding ‘3y’ and ‘5’ on both sides, we get –
$5-\dfrac{22}{29}=3y$
Taking LCM of terms $5-\dfrac{22}{29}$ , we get –
$\begin{align}
  & \dfrac{145-22}{29}=3y \\
 & \dfrac{123}{29}=3y \\
\end{align}$
By dividing both sides by ‘3’, we get –
$\dfrac{41}{29}=y$
Therefore, $x=-\dfrac{11}{29}$ and $y=\dfrac{41}{29}$
Hence, the coordinates of the points of intersection of the straight lines is $\left( -\dfrac{11}{29},\dfrac{41}{29} \right)$ .

Note: We can find the coordinates by substitution method by following a few steps. In one of the equations, express one variable in terms of the other variable. Say ‘y’ in terms of x. Substitute the value of ‘y’ and simplify the equation to find the value of ‘x’ . Then substitute the value of ‘x’ in either of the equations and solve it for ‘y’. But, this method is a long way to solve this question, so we must avoid it.