Equations with Absolute Values: Solving an exercise

To solve the equation $\left|x+4\right|=10$, we split it into two separate equations:

1. $x+4=10$

2. $x+4=-10$

For the first equation:

$x+4=10$

Subtract 4 from both sides:

$x=6$

For the second equation:

$x+4=-10$

Subtract 4 from both sides:

$x=-14$

Thus, the solutions are $x=6$ and $x=-14$.

To solve the equation $|x-1| = 6$, we will use the definition of absolute value to create two separate linear equations:

• Equation 1: $x - 1 = 6$
• Equation 2: $x - 1 = -6$

Let's solve each equation separately:

For the first equation $x - 1 = 6$:

• Add 1 to both sides of the equation:
  $x - 1 + 1 = 6 + 1$
  This simplifies to $x = 7$.

For the second equation $x - 1 = -6$:

• Add 1 to both sides of the equation:
  $x - 1 + 1 = -6 + 1$
  This simplifies to $x = -5$.

Thus, the solutions to the equation $|x-1| = 6$ are $x = -5$ and $x = 7$.

Therefore, the correct solutions are $x = -5$ and $x = 7$.

To solve the equation $\left|2x+3\right|=9$, we split it into two separate equations:

1. $2x+3=9$

2. $2x+3=-9$

For the first equation:

$2x+3=9$

Subtract 3 from both sides:

$2x=6$

Divide both sides by 2:

$x=3$

For the second equation:

$2x+3=-9$

Subtract 3 from both sides:

$2x=-12$

Divide both sides by 2:

$x=-6$

Thus, the solutions are $x=3$ and $x=-6$.

To solve the equation $\left|3x-5\right|=12$, we split it into two separate equations:

1. $3x-5=12$

2. $3x-5=-12$

For the first equation:

$3x-5=12$

Add 5 to both sides:

$3x=17$

Divide both sides by 3:

$x=5\frac{2}{3}$

For the second equation:

$3x-5=-12$

Add 5 to both sides:

$3x=-7$

Divide both sides by 3:

$x=-2\frac{1}{3}$

Thus, the solutions are $x=5\frac{2}{3}$ and $x=-2\frac{1}{3}$.