Question

How do you solve the system of equations $-9x-2y=-3\text{ and }36x+8y=13$ ?

Answer 1

Hint: To solve the system of equations $-9x-2y=-3\text{ and }36x+8y=13$ , we will compare these equations to the form ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}x={{c}_{2}}$ , where ${{a}_{1}},{{a}_{2}},{{b}_{1}},{{b}_{2}},{{c}_{1}}\text{ and }{{c}_{2}}$ are real numbers. We will check two conditions. . If we get $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ , there will be infinitely many solutions. If we obtain $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ , there will be no solution. If these two conditions are not met, then we will solve these equations using elimination, substitution or cross multiplication method.

Complete step by step solution:
We need to solve the system of equations $-9x-2y=-3\text{ and }36x+8y=13$ . Let us consider two linear equations of the form ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}x={{c}_{2}}$ , where ${{a}_{1}},{{a}_{2}},{{b}_{1}},{{b}_{2}},{{c}_{1}}\text{ and }{{c}_{2}}$ are real numbers. If we get $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ , there will be infinitely many solutions. If we obtain $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ , there will be no solution.
Now, let us check for the given pairs of equations. We can see that ${{a}_{1}}=-9,{{a}_{2}}=36,{{b}_{1}}=-2,{{b}_{2}}=8,{{c}_{1}}=-3\text{ and }{{c}_{2}}=13$ . Let us check the conditions.
$\begin{align}
  & \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{-9}{36}=\dfrac{-1}{4}...(i) \\
 & \dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{-2}{8}=\dfrac{-1}{4}...\left( ii \right) \\
 & \dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-3}{13}...\left( iii \right) \\
\end{align}$
From (i), (ii) and (iii), we can see that $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$

Hence, the given pair of linear equations does not have any solution.

Note: We can also solve this problem in an alternate way. Let us use elimination methods. For this, we have to make any one of the variable terms of one equation similar to the other so that when we add or subtract these two equations, we can eliminate one variable. Let us consider $-9x-2y=-3...(i)$ .
We have to multiply both the sides by 4. We will get
$\begin{align}
  & 4\left( -9x-2y \right)=-3\times 4 \\
 & \Rightarrow -36x-8y=-12...\left( ii \right) \\
\end{align}$
Let us consider $36x+8y=13$ as equation (iii). Now, we have to add equations (ii) and (iii).
$\begin{align}
  & -36x-8y=-12 \\
 & 36x+8y=13 \\
 & \_\_\_\_\_\_\_\_\_\_\_\_ \\
 & 0\ne 1 \\
\end{align}$
We can see that these equations cannot be solved since both the variables get cancelled and 0 cannot be equal to 1. Hence there exist no solution for the given equations. We can also use substitution and cross multiplication methods.