Question & Answer
QUESTION

Solve the following equation and find the value of y?
7y + 3 = 9

ANSWER Verified Verified
Hint: In such types of questions we take terms with y to one side of the given equation and constant terms to the other side of the equation. So, we can divide the coefficient of y with the constant term to get the value of y.

Complete step-by-step answer:
As we know that we had find the value of y in the equation,
7y + 3 = 9 …………. (1)
Now to find the value of y first, we had to take all terms of y to one side of the equation 1.
But here is only one term having variable y.
So, now we have to take all constant terms to the same side of equation 1.
So, subtract 3 to both sides of equation 1. We get,
7y + 3 – 3 = 9 – 3
7y = 6…….. (2)
Now as we know that coefficient of any variable is the constant term multiplied with it like coefficient of x in ax is a. And the coefficient of \[{{\text{x}}^2}\] in \[{\text{b}}{{\text{x}}^2}\] is b.
So, now to find the value of y in the given equation. We had to divide both sides of the equation 2 by the coefficient of y.
So, equation 2 becomes,
\[\dfrac{{{\text{7y}}}}{7} = \dfrac{6}{7}\]
y = \[\dfrac{6}{7}\]
Hence, the correct value of y in the given equation is \[\dfrac{6}{7}\].

Note: In these types of questions where we are asked to find the value of a variable and any equation is given then if the equation is linear then first, we take all terms with variable to one side of the equation and constant terms to the other side of the equation. After that we divide the constant term with the coefficient of y to get the required value of y. But if the given equation is not linear then we have to apply a formula to find a different value of y. Like if the given equation is quadratic and is \[{\text{a}}{{\text{y}}^2} + {\text{by}} + {\text{c}}\] the according to the formula different value of y will be y = \[\dfrac{{ - {\text{b}} \pm \sqrt {{{\text{b}}^2} - 4{\text{ac}}} }}{{2{\text{a}}}}\]. So, this will be the efficient way to find the value of y from the given equation.