Question

1. $x + 4 < 5(x + 3) < 7x$

Explanation:

Step1: Expand the middle - part

Expand $5(x + 3)$ to get $5x+15$. So the compound inequality becomes $x + 4<5x + 15<7x$.

Step2: Split the compound inequality

Split $x + 4<5x + 15<7x$ into two inequalities: $x + 4<5x + 15$ and $5x + 15<7x$.

Step3: Solve $x + 4<5x + 15$

Subtract $x$ from both sides: $4<4x + 15$. Then subtract 15 from both sides: $4-15<4x$, i.e., $- 11<4x$. Divide both sides by 4: $x>-\frac{11}{4}$.

Step4: Solve $5x + 15<7x$

Subtract $5x$ from both sides: $15<2x$. Divide both sides by 2: $x>\frac{15}{2}$.

Step5: Find the intersection

Since $x>-\frac{11}{4}$ and $x>\frac{15}{2}$, and $\frac{15}{2}>-\frac{11}{4}$, the solution is $x>\frac{15}{2}$.

Answer:

$x>\frac{15}{2}$