Solve the system (2/3)x − (3/4)y = 1/2 and 4(x − y) − 3(2x + y) = 17.

Step-by-step solution

1. 1

   Clear the fractions in the first equation. The denominators $3$, $4$ and $2$ have LCM $12$, so multiply every term — including the right-hand side — by $12$:

   $12\left(\tfrac23x\right)-12\left(\tfrac34y\right)=12\left(\tfrac12\right)\quad\Longrightarrow\quad 8x-9y=6$

2. 2

   Expand the second equation. Distribute both brackets, watching the sign on the $-3$:

   $4x-4y-6x-3y=17\quad\Longrightarrow\quad -2x-7y=17$

   Multiplying by $-1$ gives the tidier

   $2x+7y=-17$

3. 3

   Eliminate x. Multiply the second equation by $4$ so both have $8x$:

   $8x+28y=-68$

   and subtract the first equation $8x-9y=6$:

   $(8x+28y)-(8x-9y)=-68-6\quad\Longrightarrow\quad 37y=-74$

4. 4

   Solve for y and back-substitute.

   $y=-2$

   $2x+7(-2)=-17\quad\Longrightarrow\quad 2x-14=-17\quad\Longrightarrow\quad 2x=-3\quad\Longrightarrow\quad x=-\tfrac32$

   $\boxed{x=-\tfrac32,\quad y=-2}$

5. 5

   Check in both original equations. First: $\tfrac23\left(-\tfrac32\right)-\tfrac34(-2)=-1+\tfrac32=\tfrac12$ ✓. Second: $4\left(-\tfrac32+2\right)-3\left(-3-2\right)=4\left(\tfrac12\right)+15=2+15=17$ ✓. Checking against the originals, not the cleared versions, also confirms the multiplication by $12$ and the bracket expansion were done correctly.