Question

How do you solve the system of equations of \[6x+9y=-18\] and \[4x-9y=-42\]?

Answer 1

Hint: We can solve this question using basic linear equation concepts. We can make them a layered problem. We align the variables under each other, and then use algebra to manipulate the expressions to eliminate one unknown and solve for the other, Then we take that solved variable and use it to find the second one. Then we can find values of both the variables.

Complete step by step answer:
Given equations are
\[6x+9y=-18\]
\[4x-9y=-42\]
Here we can see that we already have equal and opposite coefficients so there is no need for any changes on equations.
If we don’t have any terms with equal coefficients then we have to make any one term to be equal by multiplying or dividing. If both of them are not having opposite signs we have to subtract otherwise we have to add.
Here we have the opposite direction so we can add them simply.
By adding both the equations we get
\[\begin{align}
  & \underline{\begin{align}
  & 6x+9y=-18 \\
 & 4x-9y=-42 \\
\end{align}} \\
 & 10x=-60 \\
\end{align}\]
Now we have to divide the equation with 10 on both sides.
\[\Rightarrow \dfrac{10x}{10}=\dfrac{-60}{10}\]
We will get
\[\Rightarrow x=-6\]
Now we will substitute this x value in the first equation given to get the y value.
Our first equation is
\[6x+9y=-18\]
Now we have to substitute x value in it
We will get
\[\Rightarrow 6\left( -6 \right)+9y=-18\]
\[\Rightarrow -36+9y=-18\]
Now grouping like terms on the side we can solve the equation.
To do this we have to add 36 to both sides of the equation.
\[\Rightarrow -36+9y+36=-18+36\]
\[\Rightarrow 9y=18\]
Now we have to divide the equation with 9 on both sides.
\[\Rightarrow \dfrac{9y}{9}=\dfrac{18}{9}\]
We will get
\[\Rightarrow y=2\]
So the x and y values by solving the equations given is
\[x=-6,y=2\]

Note: We can also do it by substitution method. We can take the value of \[9y\] from one equation and substitute it in another and then we can find both values. We can also check the answer by back substituting the values in the equation.