which of the following compound inequalities have the solution x < 3? select all that apply. a. 3x + 5 < 6…

Explanation:

Step1: Solve option A inequalities

Solve $3x + 5<6$: Subtract 5 from both sides: $3x<6 - 5$, so $3x<1$, then $x<\frac{1}{3}$. Solve $- 2x+9>3$: Subtract 9 from both sides: $-2x>3 - 9$, so $-2x>-6$. Divide both sides by - 2 and reverse the inequality sign, we get $x < 3$. The intersection of $x<\frac{1}{3}$ and $x < 3$ is $x<\frac{1}{3}$.

Step2: Solve option B inequalities

Solve $6x<5x + 3$: Subtract $5x$ from both sides, we get $x<3$. Solve $5x + 3<4x+5$: Subtract $4x$ and 3 from both sides, we get $x<2$. The intersection is $x<2$.

Step3: Solve option C inequalities

Solve $3x-5<10$: Add 5 to both sides: $3x<10 + 5$, so $3x<15$, then $x < 5$. Solve $-2x + 9>3$: Subtract 9 from both sides: $-2x>3 - 9$, so $-2x>-6$. Divide by - 2 and reverse the sign, we get $x<3$. The intersection is $x<3$.

Step4: Solve option D inequalities

Solve $3x + 5<6$: Subtract 5 from both sides: $3x<1$, so $x<\frac{1}{3}$. Solve $-2x + 9>3$: Subtract 9 from both sides: $-2x>-6$, divide by - 2 and reverse the sign, we get $x<3$. The union of $x<\frac{1}{3}$ and $x<3$ is $x<3$.

Step5: Solve option E inequalities

Solve $3x + 5<6$: Subtract 5 from both sides: $3x<1$, so $x<\frac{1}{3}$. Solve $-2x + 9<3$: Subtract 9 from both sides: $-2x<3 - 9$, so $-2x<-6$. Divide by - 2 and reverse the sign, we get $x>3$. The union is $x<\frac{1}{3}$ or $x>3$.

Answer:

A. $3x + 5<6$ and $-2x + 9>3$ B. $6x<5x + 3<4x + 5$ C. $3x-5<10$ and $-2x + 9>3$ D. $3x + 5<6$ or $-2x + 9>3$