Solve for x. 7x+39 >= 53quad AND quad16 x+15 > 31 Choose 1 answer: (A) x > 1 (B) x >= 2 (C) x <= 2 D There are no solutions (E) All values of x are solutions

Solve first inequality: Solve the first inequality 7x + 39 ≥ 53. Subtract 39 from both sides to isolate the term with x. 7x + 39 - 39 ≥ 53 - 39 7x ≥ 14Isolate x term: Divide both sides by 7 to solve for x. 7x / 7 ≥ 14 / 7 x ≥ 2Divide both sides: Solve the second inequality 16x + 15 > 31. Subtract 15 from both sides to isolate the term with x. 16x + 15 - 15 > 31 - 15 16x > 16Solve second inequality: Divide both sides by 16 to solve for x. 16x / 16 > 16 / 16 x > 1Isolate x term: Combine the solutions of both inequalities to find the common solution set. From the first inequality, we have x ≥ 2. From the second inequality, we have x > 1. The common solution set that satisfies both inequalities is x ≥ 2.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve for
x.

7x+39 >= 53quad AND
quad16 x+15 > 31
Choose 1 answer:
(A)
x > 1
(B)
x >= 2
(C)
x <= 2
D There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ $7 x+39 \geq 53 \quad$ AND \quad 16 x+15>31 $\newline$ Choose $1$ answer: $\newline$ (A) x>1 $\newline$ (B) $x \geq 2$ $\newline$ (C) $x \leq 2$ $\newline$ D There are no solutions $\newline$ (E) All values of $x$ are solutions

View step-by-step help

Home

Math Problems

Algebra 1

Solve a quadratic equation by factoring

Full solution

Q. Solve for

x

\newline

7 x+39 \geq 53 \quad

AND

\quad 16 x+15>31

\newline

Choose

1

answer:

\newline

(A)

x>1

\newline

(B)

x \geq 2

\newline

(C)

x \leq 2

\newline

D There are no solutions

\newline

(E) All values of

x

are solutions

Solve first inequality: Solve the first inequality $7x + 39 \geq 53$ . $\newline$ Subtract $39$ from both sides to isolate the term with $x$ . $\newline$ $7x + 39 - 39 \geq 53 - 39$ $\newline$ $7x \geq 14$
Isolate $x$ term: Divide both sides by $7$ to solve for $x$ . $\newline$ $\frac{$ $7$ x}{ $7$ } \geq \frac{ $14$ }{ $7$ } $\newline$ $x \geq$ $2$
Divide both sides: Solve the second inequality 16x + 15 > 31.
Subtract $15$ from both sides to isolate the term with $x$ .
16x + 15 - 15 > 31 - 15
16x > 16
Solve second inequality: Divide both sides by $16$ to solve for $x$ . $\newline$ \frac{16x}{16} > \frac{16}{16} $\newline$ x > 1
Isolate x term: Combine the solutions of both inequalities to find the common solution set. $\newline$ From the first inequality, we have $x \geq 2$ . $\newline$ From the second inequality, we have x > 1. $\newline$ The common solution set that satisfies both inequalities is $x \geq 2$ .