Question 66·Hard·Linear Equations in One Variable

Try multiplying both sides of the equation by $3-5x$ to clear the denominators. Be sure to multiply every term by $3-5x$.

After you multiply by $3-5x$, distribute the $2$ and the $-1$, then combine like terms to get a standard linear equation of the form $ax + b = cx + d$.

Once you have an equation without fractions, move all $x$-terms to one side and constants to the other, then divide to solve for $x$. Don’t forget to check that your solution doesn’t make the denominator $3-5x$ equal to zero.

In Desmos, type y1 = 4/(3-5x) + 2 to represent the left-hand side of the equation as a graph.

Type y2 = 6x/(3-5x) - 1 to represent the right-hand side. You should now see two curves (with a vertical asymptote where $3-5x = 0$).

Zoom or pan until you can see where the graphs of $y1$ and $y2$ intersect. Tap or click the intersection point; Desmos will show its coordinates. The $x$-coordinate of this intersection is the solution to the equation, as long as it does not make $3-5x$ equal to zero.

The denominator $3-5x$ appears in both fractions, so we can clear the fractions by multiplying the entire equation by $3-5x$.

First note the restriction: $3-5x \neq 0$, so $x \neq \dfrac{3}{5}$.

Multiply both sides by $3-5x$:

Now cancel the denominators in the fraction terms:

Distribute on each side:

• Left side: $2(3-5x) = 6 - 10x$, so

• Right side: $-1(3-5x) = -3 + 5x$, so

So the equation becomes

Now solve $10 - 10x = 11x - 3$ by getting all $x$-terms on one side and the constants on the other.

Add $10x$ to both sides and add $3$ to both sides:

Finally, divide both sides by $21$:

This value does not make $3-5x$ equal to zero, so it is valid. The correct answer is $\dfrac{13}{21}$.