Answer:
[tex]\textsf{B)} \quad x=-\dfrac{13}{5}[/tex]
Step-by-step explanation:
Given equation:
[tex]\dfrac{3}{4}x+\dfrac{1}{4}=-\dfrac{1}{2}x-3[/tex]
To solve the given equation, we will use algebraic operations to isolate x on one side of the equation.
Before we do this, convert the numbers on the right side of the equation to equivalent fractions with a common denominator of 4 to simplify the equation and make it easier to combine like terms during the solving process.
To convert -1/2 to an equivalent fraction with a denominator of 4, multiply both the numerator and denominator by 2.
[tex]-\dfrac{1}{2}=-\dfrac{1 \cdot 2}{2 \cdot 2}=-\dfrac{2}{4}[/tex]
We can express any whole number as a fraction by dividing it by 1. Therefore, to convert 3/1 to an equivalent fraction with a denominator of 4, multiply both the numerator and denominator by 4.
[tex]3=\dfrac{3}{1}=\dfrac{3\cdot 4}{1\cdot 4}=\dfrac{12}{4}[/tex]
Therefore the original equation becomes:
[tex]\dfrac{3}{4}x+\dfrac{1}{4}=-\dfrac{2}{4}x-\dfrac{12}{4}[/tex]
Add (2/4)x to both sides of the equation:
[tex]\begin{aligned}\dfrac{3}{4}x+\dfrac{1}{4}+\dfrac{2}{4}x&=-\dfrac{2}{4}x-\dfrac{12}{4}+\dfrac{2}{4}x\\\\\dfrac{3}{4}x+\dfrac{2}{4}x+\dfrac{1}{4}&=-\dfrac{12}{4}\end{aligned}[/tex]
Subtract 1/4 from both sides of the equation:
[tex]\begin{aligned}\dfrac{3}{4}x+\dfrac{2}{4}x+\dfrac{1}{4}-\dfrac{1}{4}&=-\dfrac{12}{4}-\dfrac{1}{4}\\\\\dfrac{3}{4}x+\dfrac{2}{4}x&=-\dfrac{12}{4}-\dfrac{1}{4}\end{aligned}[/tex]
As the fractions all have a common denominator, we can simply perform the operations on their numerators:
[tex]\begin{aligned}\dfrac{3+2}{4}x&=\dfrac{-12-1}{4}\\\\\dfrac{5}{4}x&=-\dfrac{13}{4}\end{aligned}[/tex]
To isolate x, multiply both sides of the equation by 4/5:
[tex]\begin{aligned}\dfrac{5}{4}x\cdot \dfrac{4}{5}&=-\dfrac{13}{4}\cdot \dfrac{4}{5}\\\\x&=-\dfrac{13}{5}\end{aligned}[/tex]
Therefore, the solution to the given equation is:
[tex]\Large\boxed{\boxed{x=-\dfrac{13}{5}}}[/tex]
