Solve for x. 5x-29 > -34quad OR quad2x+31 -1 (B) x -1 (D) There are no solutions (E) All values of x are solutions

Solve first inequality: Solve the first inequality 5x - 29 > -34. Add 29 to both sides of the inequality to isolate the term with x. 5x - 29 + 29 > -34 + 29 5x > -5 Now, divide both sides by 5 to solve for x. 5x/5 > -5/5 x > -1Isolate term with x: Solve the second inequality 2x + 31 -1. The second inequality gives us x -1.

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Solve for
x.

5x-29 > -34quad OR
quad2x+31 < 29
Choose 1 answer:
(A)
x < -1 or
x > -1
(B)
x < -1
(C)
x > -1
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ 5 x-29>-34 \quad OR \quad 2 x+31<29 $\newline$ Choose $1$ answer: $\newline$ (A) x<-1 or x>-1 $\newline$ (B) x<-1 $\newline$ (C) x>-1 $\newline$ (D) There are no solutions $\newline$ (E) All values of $x$ are solutions

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Math Problems

Algebra 2

Solve polynomial equations

Full solution

Q. Solve for

x

\newline

5 x-29>-34 \quad

\quad 2 x+31<29

\newline

Choose

1

answer:

\newline

(A)

x<-1

x>-1

\newline

(B)

x<-1

\newline

(C)

x>-1

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solve first inequality: Solve the first inequality 5x - 29 > -34. $\newline$ Add $29$ to both sides of the inequality to isolate the term with $x$ . $\newline$ 5x - 29 + 29 > -34 + 29 $\newline$ 5x > -5 $\newline$ Now, divide both sides by $5$ to solve for $x$ . $\newline$ \frac{5x}{5} > \frac{-5}{5} $\newline$ x > -1
Isolate term with x: Solve the second inequality 2x + 31 < 29.
Subtract $31$ from both sides of the inequality to isolate the term with $x$ .
2x + 31 - 31 < 29 - 31
2x < -2
Now, divide both sides by $2$ to solve for $x$ .
\frac{2x}{2} < \frac{-2}{2}
x < -1
Divide both sides by $5$ : Combine the solutions of both inequalities. $\newline$ The first inequality gives us x > -1. $\newline$ The second inequality gives us x < -1. $\newline$ Since these are two separate conditions connected by an ＂OR＂, the solution set includes all $x$ that satisfy either condition. $\newline$ Therefore, the solution set is x < -1 OR x > -1.