Question

How do you solve for y in the equation -2x+y = 8?

Answer 1

Hint: To solve the equation, -2x+y = 8, we will first learn how to solve such equations, then we will learn what does solving for y imply, after that, we will use certain examples to show how to operate with different algebraic expressions to simplify our answer. To solve for y, we will add both the sides with 2x and then simplify to get our answer.

Complete step by step answer:
We are given an equation, -2x+y = 8, we are asked to solve for y. Solve for y, means we have to simplify the terms in such a way that we can rewrite the equation for y.
For example, if we have x+y = 4, then solving for y means, rewriting the terms by keeping terms with yonn the left side and all the other terms on the right side. So, we can write,
x+y = 4
On subtracting x from both the sides, we get,
x+y-x = y-x
y = y-x
So, we get y = y-x, this is the equation we achieve by solving for y.
To simplify such equations, we will use algebraic operations like addition, subtraction, multiplication, and division.
If we have, 3x+2y = 8, then we will first move the terms containing x to left side, after that, we can take 2 from that side, as we are asked to solve for y.
So, we can subtract 3x from both the sides, so we get,
3x+2y-3x = 8-3x
So, we get,
2y = 8-3x
On dividing both the sides by 2, we get,
$\begin{align}
  & y=\dfrac{8}{2}-\dfrac{3x}{2} \\
 & y=4-\dfrac{3x}{2} \\
\end{align}$
Now, we are given in our question, the equation, -2x+y = 8.
We have -2x along with y, so we can add 2x on both sides in order to cancel that term. So, we get,
-2x+y+2x = 8+2x
So, we get,
y = 8+2x

Note:
 While solving such questions, always remember that variables are added to only variables of their kind. For example,
We know that 3x+2x = 5x, but we cannot 3x and 2y like, 3x+2y = 5xy or add 8 and 2x like 8+2x = 10x as doing these are incorrect steps. Only like terms can be added or subtracted together. Also, remember that we cannot add ${{x}^{2}}$ and x or any such higher power and lower powers.