Question

How do you solve \[10=-\dfrac{2}{3}(4x+5)\]?

Answer 1

Hint: We have to solve for \[x\]. We will start by multiplying \[\dfrac{3}{2}\] on either side of the equality then \[\dfrac{3}{2}\] in the RHS will get cancelled. Then because of the negative sign in the RHS instead of subtracting 5 from the expression we will add 5 on both sides of the equality and solve the equation. We have the expression in terms of \[x\], so we will solve further to get the value of \[x\].

Complete step by step solution:
According to the question, the given expression requires it to be solved for \[x\]. We will be using arithmetic operation on the expression to simplify the expression to write it in terms of \[x\].
We will start by writing the expression first, we have,
 \[10=-\dfrac{2}{3}(4x+5)\]
We will multiply both the sides of the equality by 3, we get,
\[\Rightarrow 10\times 3=\left( -\dfrac{2}{3}(4x+5) \right)\times 3\]
Solving it, we get,
\[\Rightarrow 30=\left( -2(4x+5) \right)\]
Now, we will multiply both the sides by \[\dfrac{1}{2}\] in order to remove the 2 in the RHS, we get,
\[\Rightarrow 30\times \dfrac{1}{2}=\left( -2(4x+5) \right)\times \dfrac{1}{2}\]
In the LHS, we had 30 and when it is multiplied by half we get 15 and in the RHS the 2’s got cancelled.
So, we have the expression as,
\[\Rightarrow 15=-(4x+5)\]
Opening the bracket in the RHS and due to negative sign outside there is a sign reversal of the terms within the bracket and we get,
\[\Rightarrow 15=-4x-5\]
Now, we will add 5 on both the sides and we get,
\[\Rightarrow 15+5=-4x-5+5\]
In the RHS, 5 gets cancelled and on solving we get,
\[\Rightarrow 20=-4x\]
So we have the \[x\] terms on one side and the constant on the other, solving for the value of \[x\], we get,
\[\Rightarrow -x=\dfrac{20}{4}\]
\[\Rightarrow x=-5\]

Therefore, the value of \[x=-5\].

Note: The value of the variable that we obtain from the expression can be verified if it is correct or not, by simply putting in the expression and check if \[LHS=RHS\] or not. We have the expression as,
\[10=-\dfrac{2}{3}(4x+5)\]
Taking RHS, we have,
\[-\dfrac{2}{3}(4x+5)\]
Substituting the value of \[x=-5\], we get,
\[\Rightarrow -\dfrac{2}{3}(4(-5)+5)\]
\[\Rightarrow -\dfrac{2}{3}((-20)+5)\]
\[\Rightarrow -\dfrac{2}{3}(-20+5)\]
Adding up the terms in the bracket, we have,
\[\Rightarrow -\dfrac{2}{3}(-15)\]
\[\Rightarrow -2(-5)=10=LHS\]
Since, \[LHS=RHS\], therefore we have the correct value of \[x=-5\].