NJSLA Algebra I Reasoning with Equations and Inequalities. Practice it free.

The equation means $3x - 12 = 24$ or $3x - 12 = -24$. Adding $12$ gives $3x = 36$ or $3x = -12$. Dividing by $3$ gives $x = 12$ or $x = -4$. Both solutions are $-4$ and $12\textsf{.}$

Rewrite $|6x + 12| = |6(x + 2)| = |6|\,|x + 2| = 6|x + 2|$. Since $6|x + 2| = 48$, divide by $6$ to get $|x + 2| = 8$. The expression $x + 2$ equals $8$ or $-8$; the positive value is $8\textsf{.}$

Subtract $4$ from both sides: $8x = 32$, so $x = 4$. Then $2x = 2 \cdot 4 = 8\textsf{.}$

Add 2: $3x \ge 9$. Divide by 3: $x \ge 3$, so $k = 3$.

Set the expressions equal: $6-2x=x^{2}-5x+6$, so $0=x^{2}-3x$. Then $x(x-3)=0$, and $x=0$ or $x=3$. For $x=3$, $y=6-2(3)=0$, so one intersection point is $(3,0)\textsf{.}$

Subtract the first equation from the second: $5x = 10$, so $x = 2$. From $-3x + y = 1$, $y = 7$. Then $3x - 2y = 6 - 14 = -8$.

Substitute $y = 2x + 3$ into the second equation: $3x + 2(2x + 3) = 27$, so $3x + 4x + 6 = 27$, $7x = 21$, $x = 3$, and $y = 9$.

Substitute $a = 4$: $5(4) + 3c = 41$, so $20 + 3c = 41$. Then $3c = 21$ and $c = 7$.