Solve for x. 2x+3 >= 7quad" OR "quad2x+9 > 11 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 2x + 3 ≥ 7. Subtract 3 from both sides to isolate the term with x. 2x + 3 - 3 ≥ 7 - 3 2x ≥ 4 Now, divide both sides by 2 to solve for x. 2x/2 ≥ 4/2 x ≥ 2Solve second inequality: Solve the second inequality 2x + 9 > 11. Subtract 9 from both sides to isolate the term with x. 2x + 9 - 9 > 11 - 9 2x > 2 Now, divide both sides by 2 to solve for x. 2x/2 > 2/2 x > 1Combine solutions: Combine the solutions from Step 1 and Step 2. The first inequality gives us x ≥ 2. The second inequality gives us x > 1. Since x > 1 is less restrictive than x ≥ 2, the solution that satisfies both inequalities is x ≥ 2.

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Solve for
x.

2x+3 >= 7quad" OR "quad2x+9 > 11
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 $x$ . $\newline$ 2 x+3 \geq 7 \quad \text { OR } \quad 2 x+9>11 $\newline$ 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

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Math Problems

Algebra 1

Is (x, y) a solution to the system of linear inequalities?

Full solution

Q. Solve for

x

\newline

2 x+3 \geq 7 \quad \text { OR } \quad 2 x+9>11

\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 $2x + 3 \geq 7$ . $\newline$ Subtract $3$ from both sides to isolate the term with $x$ . $\newline$ $2x + 3 - 3 \geq 7 - 3$ $\newline$ $2x \geq 4$ $\newline$ Now, divide both sides by $2$ to solve for $x$ . $\newline$ $\frac{2x}{2} \geq \frac{4}{2}$ $\newline$ $x \geq 2$
Solve second inequality: Solve the second inequality 2x + 9 > 11.
Subtract $9$ from both sides to isolate the term with $x$ .
2x + 9 - 9 > 11 - 9
2x > 2
Now, divide both sides by $2$ to solve for $x$ .
\frac{2x}{2} > \frac{2}{2}
x > 1
Combine solutions: Combine the solutions from Step $1$ and Step $2$ . $\newline$ The first inequality gives us $x \geq 2$ . $\newline$ The second inequality gives us x > 1. $\newline$ Since x > 1 is less restrictive than $x \geq 2$ , the solution that satisfies both inequalities is $x \geq 2$ .