Question

What is the solution to the following system of linear equations?

4y − 36 = −12x
3x + y = −9

Answer 1

Answer:

Given system of linear equation 4y – 36 = -12x  and 3x + y = -9 do not have any solution.

Solution:

Need to determine solution of following system of linear equations

4y – 36 = -12x

3x + y = -9

lets modify given equation in standard form

12x + 4y -36 = 0      -------(1)

3x + y + 9 = 0         -------(2)

Lets first analyze whether given system of equation is having solution or not.

$\text { If } a_{1} x+b_{1} y+c_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2}=0$ are two equation and $\left(\frac{a_{1}}{a_{2}}\right)=\left(\frac{b_{1}}{b_{2}}\right) \text { not equal to }\left(\frac{c_{1}}{c_{2}}\right)$ then the given system of equation has no solution

In our case

$\begin{array}{l}{a_{1}=12, b_{1}=4 \text { and } c_{1}=-36} \\ {a_{2}=3, b_{2}=1, \text { and } c_{2}=9}\end{array}$

$\begin{array}{l}{\frac{a_{1}}{a_{2}}=\frac{12}{3}=4} \\\\ {\frac{b_{1}}{b_{2}}=\frac{4}{1}=4} \\\\ {\frac{c_{1}}{c_{2}}=-\frac{36}{9}=-4}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=4 \text { and not equal to } \frac{c_{1}}{c_{2}}=-4$

This is condition of no solution.

Hence given system of linear equation 4y – 36 = -12x and 3x + y = -9 do not have any solution.