How do you graph and solve #4 |4-3x |= 4x + 6#?

Question

Eli Assouline · Accepted Answer

Consider that the absolute value can either be positive or negative. As a result, there will be two solutions.

Let's start with the scenario that the absolute value is positive: #4(4 - 3x) = 4x + 6# #16 - 12x = 4x + 6# #10 = 16x# #5/8 = x# Negative scenario: #4(-(4 - 3x)) = 4x + 6# #4(-4 + 3x) = 4x + 6# #-16 + 12x = 4x + 6# #-22 = -8x# #11/4 = x# The solution set is #x = 5/8 and 11/4#. Practice exercises: Solve #2 - |2x + 5| - 4x = 3x - 8#

Avery Alexander · Answer

undefined To graph and solve the equation $4\left|4-3x\right|=4x+6$, follow these steps:

1. **Split the equation into two cases**:

Case 1: $4-3x \geq 0$ which gives $4 \geq 3x$ or $x \leq \frac{4}{3}$.

Case 2: $4-3x < 0$ which gives $4 < 3x$ or $x > \frac{4}{3}$.

2. **Solve each case separately**:

For Case 1, $4 - 3x = 4 - 3x$ which simplifies to $0 = 0$. This is always true, so all $x$ values less than or equal to $\frac{4}{3}$ are solutions.

For Case 2, $4 - 3x = - (4 - 3x)$ which simplifies to $4 - 3x = -4 + 3x$. Solving for $x$, we get $x = \frac{4}{3}$.

3. **Check for extraneous solutions**:

Substitute $x = \frac{4}{3}$ back into the original equation to check if it satisfies the equation. In this case, $LHS = 4 \left|4 - 3\left(\frac{4}{3}\right)\right| = 4 \times 1 = 4$, and $RHS = 4 \times \frac{4}{3} + 6 = \frac{16}{3} + 6 = \frac{34}{3} \neq 4$. So, $x = \frac{4}{3}$ is not a solution.

4. **Graph the solutions**:

On a number line, mark $\frac{4}{3}$ with an open circle to indicate it's not included, then shade the region to the left to represent $x \leq \frac{4}{3}$. This is the solution set for the absolute value equation.