Question

1. (4x + 1 < -6(x + 2) leq 7x) (4x + 1 < -6x - 18 leq 7x)

Explanation:

Step1: Split the compound - inequality

Split $4x + 1<-6(x + 2)\leq7x$ into two inequalities: $4x + 1<-6(x + 2)$ and $-6(x + 2)\leq7x$.

Step2: Solve $4x + 1<-6(x + 2)$

Expand the right - hand side: $4x + 1<-6x-12$. Add $6x$ to both sides: $4x+6x + 1<-12$, which simplifies to $10x+1<-12$. Then subtract 1 from both sides: $10x<-13$, so $x<-\frac{13}{10}$.

Step3: Solve $-6(x + 2)\leq7x$

Expand the left - hand side: $-6x-12\leq7x$. Add $6x$ to both sides: $-12\leq7x + 6x$, which simplifies to $-12\leq13x$. Then divide both sides by 13: $x\geq-\frac{12}{13}$.
Since $-\frac{12}{13}>-\frac{13}{10}$ is false, there is no solution for this compound inequality.

Answer:

No solution