Question

1. solve $16(t - 1) + 10 > 8(t + 2) + 4(t - 1) + 4t$.

a. $\circ$
b. $\circ t < 1$
c. $\circ t > -1$
d. $\circ t > 1$

1. solve $-4(10 - n) - 6n > 8(n + 3) - 7n + 2$.

a. $\circ n > -22$
b. $\circ n < -6$
c. $\circ n < -22$
d. $\circ n > -6$

1. solve $3(m - 2) + 5m \leq 2(m + 1) + 4$

a. $\circ m \geq 2$
b. $\circ m \leq 0$
c. $\circ m \geq 0$
d. $\circ m < 2$

Explanation:

Problem 2

Step 1: Expand both sides

Expand \(16(t - 1)+10\): \(16t-16 + 10=16t-6\)
Expand \(8(t + 2)+4(t - 1)+4t\): \(8t + 16+4t-4 + 4t=16t + 12\)

Step 2: Solve the inequality

We have \(16t-6>16t + 12\). Subtract \(16t\) from both sides: \(-6>12\), which is false. Wait, maybe a typo? Wait, original problem: \(16(t - 1)+10>8(t + 2)+4(t - 1)+4t\). Wait, let's re - expand:

Left side: \(16t-16 + 10=16t - 6\)
Right side: \(8t+16 + 4t-4 + 4t=16t + 12\)
\(16t-6>16t + 12\) → \(-6>12\), no solution? But option A is "∅"? Wait, maybe I misread. Wait, maybe the problem is \(16(t - 1)+10\geq8(t + 2)+4(t - 1)+4t\)? No, as per given. Wait, maybe the original problem has a typo, but if we proceed, the inequality \(16t-6>16t + 12\) simplifies to \(-6>12\), which is never true, so the solution set is empty. So option A.

Step 1: Expand both sides

Expand \(-4(10 - n)-6n\): \(-40 + 4n-6n=-40 - 2n\)
Expand \(8(n + 3)-7n + 2\): \(8n+24-7n + 2=n + 26\)

Step 2: Solve the inequality

We have \(-40-2n>n + 26\). Add \(2n\) to both sides: \(-40>3n + 26\). Subtract 26 from both sides: \(-66>3n\). Divide by 3: \(n<-22\)

Step 1: Expand both sides

Expand \(3(m - 2)+5m\): \(3m-6 + 5m=8m-6\)
Expand \(2(m + 1)+4\): \(2m+2 + 4=2m + 6\)

Step 2: Solve the inequality

We have \(8m-6\leq2m + 6\). Subtract \(2m\) from both sides: \(6m-6\leq6\). Add 6 to both sides: \(6m\leq12\). Divide by 6: \(m\leq2\)? Wait, no, wait the original problem: \(3(m - 2)+5m\leq2(m + 1)+4\). Wait, my expansion:

Left: \(3m-6 + 5m=8m-6\)
Right: \(2m + 2+4=2m + 6\)
\(8m-6\leq2m + 6\)
\(8m-2m\leq6 + 6\)
\(6m\leq12\)
\(m\leq2\)? But options: A. \(m\geq2\), B. \(m\leq0\), C. \(m\geq0\), D. \(m < 2\). Wait, maybe I misread the problem. Wait, the problem is \(3(m - 2)+5m\leq2(m + 1)+4\)? Wait, let's re - calculate:

\(3m-6 + 5m=8m-6\)
\(2m+2 + 4=2m + 6\)
\(8m-6\leq2m + 6\)
\(8m-2m\leq6 + 6\)
\(6m\leq12\)
\(m\leq2\). But option D is \(m < 2\), A is \(m\geq2\), B \(m\leq0\), C \(m\geq0\). Wait, maybe a typo in the problem. Wait, if the problem is \(3(m - 2)+5m\leq2(m + 1)+4\), then \(m\leq2\). But if the problem was \(3(m - 2)+5m\geq2(m + 1)+4\), then \(8m-6\geq2m + 6\) → \(6m\geq12\) → \(m\geq2\) (option A). Maybe I misread the inequality sign. The original problem: "Solve \(3(m - 2)+5m\leq2(m + 1)+4\)"? Wait, the user's image: "Solve \(3(m - 2)+5m\leq2(m + 1)+4\)". Wait, but with the given options, maybe there is a mistake. Wait, let's check again:

\(3(m - 2)+5m=3m-6 + 5m=8m-6\)
\(2(m + 1)+4=2m+2 + 4=2m + 6\)
\(8m-6\leq2m + 6\)
\(6m\leq12\)
\(m\leq2\). But the options are A. \(m\geq2\), B. \(m\leq0\), C. \(m\geq0\), D. \(m < 2\). None of these is \(m\leq2\) exactly, but D is \(m < 2\), maybe a typo in the problem (maybe the inequality is \(3(m - 2)+5m<2(m + 1)+4\)). Let's solve \(3(m - 2)+5m<2(m + 1)+4\):

\(8m-6<2m + 6\)
\(6m<12\)
\(m < 2\), which is option D.

Answer:

A. ∅

Problem 3