Question 39·Hard·Linear Equations in One Variable

After distributing, carefully subtract the second expression on the left. Be sure to distribute the minus sign as well, then combine $x$-terms and constants.

Once you simplify the left side, you should get an equation where the variable terms on both sides look similar. Try moving all $x$-terms to one side and see what kind of statement you are left with.

After you eliminate $x$, check whether the remaining equation (with just numbers) is always true, never true, or sometimes true. Connect that to how many solutions the original equation has.

In Desmos, enter the left side as one function and the right side as another:

• In the first line, type: y = (4/9)(9x - 15) - (1/3)(6x - 10)
• In the second line, type: y = 2x - 5.

Look at the graph and see whether the two lines intersect. The number of intersection points of the two graphs is the same as the number of solutions to the equation.

Start by distributing each fraction to the terms inside its parentheses.

• For the first term:

  • $\frac{4}{9} \cdot 9x = 4x$
  • $\frac{4}{9} \cdot (-15) = -\frac{60}{9} = -\frac{20}{3}$
  • So $\frac{4}{9}(9x - 15) = 4x - \frac{20}{3}$.

• For the second term:

  • $\frac{1}{3} \cdot 6x = 2x$
  • $\frac{1}{3} \cdot (-10) = -\frac{10}{3}$
  • So $\frac{1}{3}(6x - 10) = 2x - \frac{10}{3}$.

The equation becomes:

Carefully subtract the second parentheses on the left-hand side:

Combine like terms:

• $4x - 2x = 2x$
• $-\tfrac{20}{3} + \tfrac{10}{3} = -\tfrac{10}{3}$

So the left-hand side simplifies to $2x - \tfrac{10}{3}$, and the equation is now

Subtract $2x$ from both sides to see what remains:

This simplifies to

Now observe whether this statement can ever be true.

The equation $-\tfrac{10}{3} = -5$ is false, because $-\tfrac{10}{3} = -3.\overline{3}$ and $-5$ are not equal. This means there is no value of $x$ that makes the original equation true. Therefore, the equation has zero solutions.