Question

for problems 2 - 5, solve and graph the solution to the inequality.

1. $3x + 5 < 8$
2. $-4x + 3 \leq 7$
3. $6x + 4 \geq 10$
4. $-5x + 2 > 7$

Explanation:

Step1: Solve \(3x + 5<8\)

Subtract \(5\) from both sides: \(3x+5 - 5<8 - 5\), so \(3x<3\).
Divide both sides by \(3\): \(x < 1\).
On the number - line, we use an open circle at \(x = 1\) (since \(x
eq1\)) and shade to the left.

Step2: Solve \(-4x + 3\leq7\)

Subtract \(3\) from both sides: \(-4x+3 - 3\leq7 - 3\), so \(-4x\leq4\).
Divide both sides by \(- 4\) and reverse the inequality sign: \(x\geq - 1\).
On the number - line, we use a closed circle at \(x=-1\) (since \(x = - 1\) is included) and shade to the right.

Step3: Solve \(6x + 4\geq10\)

Subtract \(4\) from both sides: \(6x+4 - 4\geq10 - 4\), so \(6x\geq6\).
Divide both sides by \(6\): \(x\geq1\).
On the number - line, we use a closed circle at \(x = 1\) (since \(x = 1\) is included) and shade to the right.

Step4: Solve \(-5x + 2>7\)

Subtract \(2\) from both sides: \(-5x+2 - 2>7 - 2\), so \(-5x>5\).
Divide both sides by \(-5\) and reverse the inequality sign: \(x<-1\).
On the number - line, we use an open circle at \(x=-1\) (since \(x
eq - 1\)) and shade to the left.

Answer:

1. \(x < 1\); Open circle at \(x = 1\), shade left.
2. \(x\geq - 1\); Closed circle at \(x=-1\), shade right.
3. \(x\geq1\); Closed circle at \(x = 1\), shade right.
4. \(x<-1\); Open circle at \(x=-1\), shade left.