Question

How do you solve for y in $5x + 4y = 10?$

Answer 1

Hint: Here we need to solve for the variable y in the equation $5x + 4y = 10$. i.e. we need to isolate the term y on the L.H.S. We make use of some mathematical operations to do this. Firstly, we try to keep the terms containing y on L.H.S. and take all the other terms to R.H.S. Then in R.H.S. simplify the terms to obtain the solution and then we obtain the value for the variable y.

Complete step by step solution:
Given an equation $5x + 4y = 10$ ……(1)
Here it is mentioned that we need to solve for the variable y. i.e. we need to find the value of y.
We try to do this by taking the terms which do not contain the variable y to the other side and simplify it.
Firstly, we keep the terms only containing y in L.H.S. i.e. here to keep the term $4y$ in L.H.S.
Move all terms which do not contain the variable y to R.H.S.
So we subtract $5x$on both sides of the equation (1) we get,
$ \Rightarrow 5x - 5x + 4y = 10 - 5x$
Now combining the like terms on L.H.S. $5x - 5x = 0$
Hence, the above equation becomes,
$ \Rightarrow 0 + 4y = 10 - 5x$
$ \Rightarrow 4y = 10 - 5x$
Since we need only the variable y on the left hand side, we transfer the coefficient of y to the other side.
Now dividing the whole equation by 4, we get,
$ \Rightarrow \dfrac{{4y}}{4} = \dfrac{{10 - 5x}}{4}$
$ \Rightarrow y = \dfrac{{10 - 5x}}{4}$
Simplifying the terms on the R.H.S. we get,
$ \Rightarrow y = \dfrac{{10}}{4} - \dfrac{{5x}}{4}$
$ \Rightarrow y = \dfrac{5}{2} - \dfrac{{5x}}{4}$

Hence the solution for y in the equation $5x + 4y = 10$ is $y = \dfrac{5}{2} - \dfrac{{5x}}{4}$.

Note :
Since it is mentioned to solve for y we found the value for the variable y. In a similar manner we can also find the solution for the variable x.
We can verify whether the solution we obtained is correct by substituting back the value of y in the given equation. If the equation is satisfied then the obtained value of is the required solution for a given problem.
It is important to know the following basic facts.
An equation remains unchanged or undisturbed if it satisfies the following conditions.
(1) If L.H.S. and R.H.S. are interchanged.
(2) If the same number is added on both sides of the equation.
(3) If the same number is subtracted on both sides of the equation.
(4) When both L.H.S. and R.H.S. are multiplied by the same number.
(5) When both L.H.S. and R.H.S. are divided by the same number.