Question

solve the equation.

1. $x - 7 = 15$
2. $\frac{2}{3}x + 5 = 3$
3. $2|x - 3| - 5 = 7$
4. $|2x - 19| = 4x + 1$
5. $-2 + 5x - 7 = 3x - 9 + 2x$
6. $\frac{1}{3}(6x + 12) - 2(x - 7) = 19$

Explanation:

Step1: Solve $x-7=15$

Add 7 to both sides:
$x = 15 + 7$

Step2: Solve $\frac{2}{3}x+5=3$

Subtract 5, multiply by $\frac{3}{2}$:
$\frac{2}{3}x = 3 - 5 = -2$
$x = -2 \times \frac{3}{2}$

Step3: Solve $2|x-3|-5=7$

Isolate absolute value, split cases:
$2|x-3| = 7 + 5 = 12$
$|x-3| = 6$
Case1: $x-3=6 \implies x=6+3$
Case2: $x-3=-6 \implies x=-6+3$

Step4: Solve $|2x-19|=4x+1$

Split cases, check validity:
Case1: $2x-19=4x+1$
$-2x = 20 \implies x=-10$
Check: $4(-10)+1=-39 < 0$, invalid (absolute value ≥0)
Case2: $2x-19=-(4x+1)$
$2x-19=-4x-1$
$6x=18 \implies x=3$
Check: $4(3)+1=13 ≥0$, valid

Step5: Solve $-2+5x-7=3x-9+2x$

Simplify both sides:
$5x - 9 = 5x - 9$
$0=0$

Step6: Solve $\frac{1}{3}(6x+12)-2(x-7)=19$

Simplify left side:
$2x + 4 - 2x + 14 = 19$
$18 = 19$

Answer:

1. $x=22$
2. $x=-3$
3. $x=9$ or $x=-3$
4. $x=3$
5. All real numbers (identity equation)
6. No solution (contradiction)