Question

How do you solve $0x-8y=-16$ and $-8x+2y=36$ ?

Answer 1

Hint: We are given two equations as $0x-8y=-16$ and $-8x+2y=36$ .
We have to solve these, to do so we will first find what type of equation we have then we will find what are ways to solve such kind of problems, we will use the elimination method to solve our problem. We will also try substitution methods to get more than 1 method to solve our problem.
In eliminatory, we eliminate ‘x’ and find the value of ‘y’ first then solve for ‘x’ after getting the value of ‘y’.

Complete step by step answer:
We are given $0x-8y=-16$ and $-8x+2y=36$ .
It is an equation in variable ‘x’ and ‘y’, and each variable has the highest power as 1, so they are linear equations in two variables.
There are more than one way to solve the problem: have a linear equation in 2 variables.
Some of the methods are –
1, Eliminatory method
2, Cross multiplication method
3, Graphical method
4, Substitution method
Each method criteria is different to find a solution but each method will give us the same solution.
Now as we see that in our equation –
$0x-8y=-16$ …………………………….(1)
And $-8x+2y=36$………………………….. (2)
Variable ‘x’ has coefficient 0 in equation (1) so we will use elimination method, we will eliminate ‘x’.
We will multiply equation (2) by 0. So we get –
$0x+0y=0$ …………………………. (3)
Now, we add equation (1) and equation (3).
So, we have –
\[\begin{align}
  & 0x-8y=-16 \\
 & 0x+0y=0 \\
 & \text{ }-8y=-16 \\
\end{align}\]
So, we get –
$\Rightarrow -8y=-16$ , now we simplify by dividing both sides by -8.
So, we get –
$y=+2$
So, the value of ‘y’ is 2.
Now we put $y=2$ in equation (2) to get the value of ‘x’.
Equation (2) is $-8x+2y=36$ , putting $y=2$ is here we get –
$\Rightarrow -8x+2\left( 2 \right)=36$ .
By soling, we get –
$-8x+4=36\Rightarrow -8x=36-4$ .
Solving for ‘x’, we get –
$\Rightarrow -8x=32$ .
By dividing both sides by -8, we get –
$\Rightarrow x=-4$ .
Hence $x=-4$ and $y=2$ is the solution to our problem.

Note:
 We can use other methods also, we can try using a substitution method, in this we find the value of 1 variable in terms of another variable using one equation and then substitute it in another equation.
Our first equation is $0x-8y=-16$
So, we get –
$\begin{align}
  & \Rightarrow -8y=-16-0x \\
 & \Rightarrow y=\dfrac{-16-0x}{-8} \\
\end{align}$ .
We substitute ‘y’ as $\dfrac{-16-0x}{-8}$ in equation (2).
We get –
$-8x+2\left( \dfrac{-16-0x}{-8} \right)=36$
Now we simplify, we open bracket and get –
$\Rightarrow -8x+2\times \left( \dfrac{-16}{-8} \right)-2\left( 0x \right)=36$
By solving further, we get –
$\Rightarrow -8x+4=36$ (as $\dfrac{-16}{-8}=2$ and $2\times 0x=0$ )
Now we subtract ‘4’ on both sides, so
$\Rightarrow -8x=32$
By dividing both sides by 8, we get –
$x=-4$ .
As we get $x=-4$ , we put $x=-4$ in $y=\dfrac{-16-0\left( x \right)}{-8}$ .
We will get $y=\dfrac{-16-0\left( -4 \right)}{8}$ .
$\Rightarrow y=\dfrac{-16}{-8}$ (as $0\times \left( -4 \right)=0$
So, we get $y=2$ .
Hence $y=2$ and $x=-4$ in the solution both methods gave us the same solution.