Solve ((4r−3)/(6r+1) = (2r−1)/(3r+4))

Question

ScanSolve AI Tutor · Accepted Answer

Here is the solution for question 3 from the image.

3. Solve $\frac{4r-3}{6r+1} = \frac{2r-1}{3r+4}$

Cross-multiply the terms.
$(4r-3)(3r+4) = (2r-1)(6r+1)$
Expand both sides of the equation using the FOIL method.
Left side:
$(4r)(3r) + (4r)(4) + (-3)(3r) + (-3)(4)$ $12r^2 + 16r - 9r - 12$ $12r^2 + 7r - 12$
Right side:
$(2r)(6r) + (2r)(1) + (-1)(6r) + (-1)(1)$ $12r^2 + 2r - 6r - 1$ $12r^2 - 4r - 1$
Set the expanded expressions equal to each other.
$12r^2 + 7r - 12 = 12r^2 - 4r - 1$
Move all terms to one side to solve for $r$ .
Subtract $12r^2$ from both sides.
$7r - 12 = -4r - 1$
Add $4r$ to both sides.
$7r + 4r - 12 = -1$ $11r - 12 = -1$
Add 12 to both sides.
$11r = -1 + 12$ $11r = 11$
Divide by 11.
$r = \frac{11}{11}$ $r = 1$
Check for extraneous solutions (values of $r$ that make the denominators zero).
For $6r+1$ : $6(1)+1 = 7 \neq 0$
For $3r+4$ : $3(1)+4 = 7 \neq 0$
Since the denominators are not zero, $r=1$ is a valid solution.

The final answer for 3 is $\boxed{r=1}$ .

Solve \((4r-3)/(6r+1) = (2r−1)/(3r+4)\)