what is the solution to $\frac{3}{4}(x + 8)>\frac{1}{2}(2x + 10)$?\n$(-infty,-4)$\n$(-4,infty)$\n$(-infty,4)$…
what is the solution to $\frac{3}{4}(x + 8)>\frac{1}{2}(2x + 10)$?\n$(-infty,-4)$\n$(-4,infty)$\n$(-infty,4)$\n$(4,infty)$
Answer
Explanation:
Step1: Expand both sides
$\frac{3}{4}(x + 8)=\frac{3}{4}x+6$ and $\frac{1}{2}(2x + 10)=x + 5$. So the inequality becomes $\frac{3}{4}x+6>x + 5$.
Step2: Move x - terms to one side
Subtract $\frac{3}{4}x$ from both sides: $6>x-\frac{3}{4}x + 5$. Simplify to get $6>\frac{1}{4}x+5$.
Step3: Isolate x
Subtract 5 from both sides: $1>\frac{1}{4}x$. Multiply both sides by 4: $4>x$, which can be written as $x < 4$ or $(-\infty,4)$.
Answer:
$(-\infty,4)$