Question

Solve for $x$ and $y$: $11x+15y+23=0,7x-2y-20=0$

Answer 1

Hint: Here we have been given two equations and we have to find the value of the two unknown variables. We will use cross multiplication to solve the equation. Firstly we will write the general form of the equation and compare it with the equation given then we will substitute the values in the formula and simplify it to get our desired answer.

Complete step by step answer:
We have been given two equations as,
$11x+15y+23=0$…..$\left( 1 \right)$
$7x-2y-20=0$…..$\left( 2 \right)$
The general forms of equations are,
$ax+by+c=0$ …..$\left( 3 \right)$
$px+qy+r=0$……$\left( 4 \right)$
Formula for cross multiplication of above two equations is,
$\dfrac{x}{br-cq}=\dfrac{y}{cp-ar}=\dfrac{1}{aq-bp}$
$\Rightarrow x=\dfrac{br-cq}{aq-bp}$ , $y=\dfrac{cp-ar}{aq-bp}$ ……$\left( 5 \right)$
Comparing equation (1) by (3) and (2) by (4) we get,
$a=11,b=15,c=23$
$p=7,q=-2,r=-20$
Substituting the above two values in equation (5) we get,
$x=\dfrac{\left( 15\times -20 \right)-\left( 23\times -2 \right)}{\left( 11\times -2 \right)-\left( 15\times 7 \right)}$ , $y=\dfrac{\left( 23\times 7 \right)-\left( 11\times -20 \right)}{\left( 11\times -2 \right)-\left( 15\times 7 \right)}$
Simplifying them one-by-one we get,
$\Rightarrow x=\dfrac{-300-\left( -46 \right)}{-22-105}$
$\Rightarrow x=\dfrac{-300+46}{-127}$
So we get,
$\Rightarrow x=\dfrac{-254}{-127}$
$\Rightarrow x=2$
Next we will solve $y$ value as,
$\Rightarrow y=\dfrac{161-\left( -220 \right)}{-22-105}$
$\Rightarrow y=\dfrac{161+220}{-127}$
So we get,
$\Rightarrow y=\dfrac{381}{-127}$
$\Rightarrow y=-3$
So we get $x=2$ and $y=-3$
Hence on solving equation $11x+15y+23=0,7x-2y-20=0$ we get the values as $x=2$ and $y=-3$ .

Note:
We can solve this question using the 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. As the coefficients in the equations are little big so using elimination methods will need more calculation and there is a chance of error.