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Solve for
x.

5x-19 <= 1quad OR
quad-4x+3 < -6
Choose 1 answer:
(A)
x >= 4
(B)
(9)/(4) < x <= 4
(C)
x > -(9)/(4)
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ $5 x-19 \leq 1 \quad$ OR \quad-4 x+3<-6 $\newline$ Choose $1$ answer: $\newline$ (A) $x \geq 4$ $\newline$ (B) $\frac{9}{4}<x 4="" \leq=""$ $\newline$ (c)="" $="" x="">-\frac{9}{4}$ $\newline$ (D) There are no solutions $\newline$ (E) All values of $x$ are solutions

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

Full solution

Q. Solve for

x

.

\newline

5 x-19 \leq 1 \quad

OR

\quad-4 x+3<-6

\newline

Choose

1

answer:

\newline

(A)

x \geq 4

\newline

(B)

\frac{9}{4}<x \leq 4

\newline

(C)

x>-\frac{9}{4}

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solve first inequality: Solve the first inequality $5x - 19 \leq 1$ . $\newline$ Add $19$ to both sides of the inequality to isolate the term with $x$ on one side. $\newline$ $5x - 19 + 19 \leq 1 + 19$ $\newline$ $5x \leq 20$ $\newline$ Now, divide both sides by $5$ to solve for $x$ . $\newline$ $x \leq \frac{20}{5}$ $\newline$ $x \leq 4$
Isolate x term: Solve the second inequality -4x + 3 < -6. $\newline$ Subtract $3$ from both sides of the inequality to isolate the term with $x$ on one side. $\newline$ -4x + 3 - 3 < -6 - 3 $\newline$ -4x < -9 $\newline$ Now, divide both sides by $-4$ , remembering to reverse the inequality sign because we are dividing by a negative number. $\newline$ x > \frac{-9}{-4} $\newline$ x > \frac{9}{4}
Solve second inequality: Combine the solutions from Step $1$ and Step $2$ to find the solution set for $x$ . $\newline$ From Step $1$ , we have $x \leq 4$ . $\newline$ From Step $2$ , we have x > \frac{9}{4}. $\newline$ The solution set is the intersection of these two inequalities, which is \frac{9}{4} < x \leq 4.

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