Question

Solve ${\left( {y - 3} \right)^2} = 4y - 12$.

Answer 1

Hint: The above given expression is an example of a two step equation. In order to solve it we need to manipulate the given equation in such a way that we should get $y$ by itself. In order to get $x$ by itself we can perform any arithmetic operations on both LHS and RHS equally at the same time such that the equality of the given equation doesn’t change.

Complete step-by-step solution:
Given
${\left( {y - 3} \right)^2} = 4y - 12..................................\left( i \right)$
Now in order to solve the given equation we need to solve for $y$.
Such that we have to manipulate the given equation in terms of only $y$, which can be achieved by performing different arithmetic operations on both LHS and RHS equally.
So to isolate the $y$ term from equation (i) first we have to take 4 common from both the terms in the RHS.
Such that:
\[
   \Rightarrow {\left( {y - 3} \right)^2} = 4y - 12 \\
   \Rightarrow {\left( {y - 3} \right)^2} = 4\left( {y - 3} \right).......................................\left( {ii} \right) \\
 \]
Now we can see that $\left( {y - 3} \right)$ is common to both RHS and LHS such that one $\left( {y - 3} \right)$ can be cancelled.
So we get:
\[
   \Rightarrow {\left( {y - 3} \right)^2} = 4\left( {y - 3} \right) \\
   \Rightarrow \left( {y - 3} \right) = 4..........................\left( {iii} \right) \\
 \]
Now to isolate the term $y$ we have $ + 3$ to both LHS and RHS, since adding $ + 3$ to the LHS will cancel the term $ - 3$.
Such that:
\[
   \Rightarrow \left( {y - 3} \right) = 4 \\
   \Rightarrow y - 3 + 3 = 4 + 3 \\
   \Rightarrow y = 7..............................\left( {iv} \right) \\
 \]

Therefore on solving ${\left( {y - 3} \right)^2} = 4y - 12$ we get\[y = 7\].
Note: A two-step equation is an algebraic equation which can be solved in two steps. The equation is said to be true when we find the value of the variable which makes the equation true. We can also check if the value of the variable that we got is true or not by substituting the value of the variable back into the equation and checking whether it satisfies the given equation or not.