Solve for x. 3x-91 > -87quad AND quad17 x-16 > 18 Choose 1 answer: (A) x > 2 (B) x > (4)/(3) (C) (4)/(3) < x < 2 D There are no solutions (E) All values of x are solutions

Solve first inequality: Solve the first inequality 3x - 91 > -87. Add 91 to both sides to isolate the term with x. 3x - 91 + 91 > -87 + 91 3x > 4 Now, divide both sides by 3 to solve for x. 3x/3 > 4/3 x > 4/3Solve second inequality: Solve the second inequality 17x - 16 > 18. Add 16 to both sides to isolate the term with x. 17x - 16 + 16 > 18 + 16 17x > 34 Now, divide both sides by 17 to solve for x. 17x/17 > 34/17 x > 2Combine solutions: Combine the solutions from Step 1 and Step 2. We have two inequalities: x > 4/3 x > 2 Since x must be greater than both 4/3 and 2, and 2 is greater than 4/3, the solution set is determined by the more restrictive inequality, which is x > 2.

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Solve for
x.

3x-91 > -87quad AND
quad17 x-16 > 18
Choose 1 answer:
(A)
x > 2
(B)
x > (4)/(3)
(C)
(4)/(3) < x < 2
D There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ 3 x-91>-87 \quad AND \quad 17 x-16>18 $\newline$ Choose $1$ answer: $\newline$ (A) x>2 $\newline$ (B) x>\frac{4}{3} $\newline$ (C) \( \frac{4}{3}

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Math Problems

Algebra 1

Solve a quadratic equation by factoring

Full solution

Q. Solve for

x

\newline

3 x-91>-87 \quad

AND

\quad 17 x-16>18

\newline

Choose

1

answer:

\newline

(A)

x>2

\newline

(B)

x>\frac{4}{3}

\newline

(C)

\frac{4}{3}<x<2

\newline

D There are no solutions

\newline

(E) All values of

x

are solutions

Solve first inequality: Solve the first inequality 3x - 91 > -87. $\newline$ Add $91$ to both sides to isolate the term with $x$ . $\newline$ 3x - 91 + 91 > -87 + 91 $\newline$ 3x > 4 $\newline$ Now, divide both sides by $3$ to solve for $x$ . $\newline$ \frac{3x}{3} > \frac{4}{3} $\newline$ x > \frac{4}{3}
Solve second inequality: Solve the second inequality 17x - 16 > 18. $\newline$ Add $16$ to both sides to isolate the term with $x$ . $\newline$ 17x - 16 + 16 > 18 + 16 $\newline$ 17x > 34 $\newline$ Now, divide both sides by $17$ to solve for $x$ . $\newline$ \frac{17x}{17} > \frac{34}{17} $\newline$ x > 2
Combine solutions: Combine the solutions from Step $1$ and Step $2$ . $\newline$ We have two inequalities: $\newline$ x > \frac{4}{3} $\newline$ x > 2 $\newline$ Since $x$ must be greater than both $\frac{4}{3}$ and $2$ , and $2$ is greater than $\frac{4}{3}$ , the solution set is determined by the more restrictive inequality, which is x > 2.