Solve for x. 11 x+4 -25 Choose 1 answer: (A) -(3)/(2) -(3)/(2) (C) x < 1 (D) There are no solutions (E) All values of x are solutions

Solve first inequality: Solve the first inequality 11x + 4 -25. Add 7 to both sides to isolate the term with x. 12x - 7 + 7 > -25 + 7 12x > -18 Now, divide both sides by 12 to solve for x. 12x / 12 > -18 / 12 x > -1.5 or x > -(3/2)Combine solutions: Combine the solutions from Step 1 and Step 2. The first inequality gives us x -(3/2). Since we are looking for values of x that satisfy either inequality (due to the ＂OR＂ condition), we combine the two solution sets. This means x can be any value greater than -(3/2) and less than 1, or x can be any value greater than -(3/2) alone.Determine correct answer choice: Determine the correct answer choice. Looking at the answer choices, we need to find the one that matches our combined solution set. (A) -(3/2) -(3/2) is correct because it includes all values of x that are greater than -(3/2), which is the solution to the second inequality and is not limited by the first inequality. (C) x < 1 is not correct because it does not include values of x that are less than -(3/2). (D) There are no solutions is not correct because we found solutions for x. (E) All values of x are solutions is not correct because x cannot be equal to or greater than 1 according to the first inequality.

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Solve for
x.

11 x+4 < 15quad OR
quad12 x-7 > -25
Choose 1 answer:
(A)
-(3)/(2) < x < 1
(B)
x > -(3)/(2)
(C)
x < 1
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ 11 x+4<15 \quad OR \quad 12 x-7>-25 $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{3}{2}<x<1$ $\newline$ (b)="" $="" x="">-\frac{3}{2}$ $\newline$ (C) x<1 $\newline$ (D) There are no solutions $\newline$ (E) All values of $x$ are solutions

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Math Problems

Algebra 1

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

Full solution

Q. Solve for

x

\newline

11 x+4<15 \quad

\quad 12 x-7>-25

\newline

Choose

1

answer:

\newline

(A)

-\frac{3}{2}<x<1

\newline

(B)

x>-\frac{3}{2}

\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 11x + 4 < 15.
Subtract $4$ from both sides to isolate the term with $x$ .
11x + 4 - 4 < 15 - 4
11x < 11
Now, divide both sides by $11$ to solve for $x$ .
\frac{11x}{11} < \frac{11}{11}
x < 1
Solve second inequality: Solve the second inequality 12x - 7 > -25. $\newline$ Add $7$ to both sides to isolate the term with $x$ . $\newline$ 12x - 7 + 7 > -25 + 7 $\newline$ 12x > -18 $\newline$ Now, divide both sides by $12$ to solve for $x$ . $\newline$ \frac{12x}{12} > \frac{-18}{12} $\newline$ x > -1.5 or x > -\left(\frac{3}{2}\right)
Combine solutions: Combine the solutions from Step $1$ and Step $2$ . $\newline$ The first inequality gives us x < 1. $\newline$ The second inequality gives us x > -\left(\frac{3}{2}\right). $\newline$ Since we are looking for values of $x$ that satisfy either inequality (due to the ＂OR＂ condition), we combine the two solution sets. $\newline$ This means $x$ can be any value greater than $-\left(\frac{3}{2}\right)$ and less than $1$ , or $x$ can be any value greater than $-\left(\frac{3}{2}\right)$ alone.
Determine correct answer choice: Determine the correct answer choice. $\newline$ Looking at the answer choices, we need to find the one that matches our combined solution set. $\newline$ (A) -(\frac{3}{2}) < x < 1 is not correct because it does not include values of $x$ that are greater than $1$ . $\newline$ (B) x > -(\frac{3}{2}) is correct because it includes all values of $x$ that are greater than $-(\frac{3}{2})$ , which is the solution to the second inequality and is not limited by the first inequality. $\newline$ (C) x < 1 is not correct because it does not include values of $x$ that are less than $-(\frac{3}{2})$ . $\newline$ (D) There are no solutions is not correct because we found solutions for $x$ . $\newline$ (E) All values of $x$ are solutions is not correct because $x$ cannot be equal to or greater than $1$ according to the first inequality.