What values of x and y satisfy `-(13)/(3)x −2y = 7` and `3x + (2)/(5) y = 5` simultaneously? a. x = 3 and y = −10 b. x = −3 and y = −3 c. x = 1 and y = 5 d. x = 3 and y = 10

I like to do this using the substitution method! First, you have to get a variable alone in one of the two equations. Then, you plug in the statement for that variable in the other equation and solve for the other variable. Finally, plug that variable into one of the equations and solve for the missing variable. It looks something like this:

Solve for y in the first equation:

\frac{-13}{3}x-2y= 7 \\ (\frac{-13}{3}x-2y)+\frac{13}{3}x= (7) +\frac{13}{3}x \\ -2y = \frac{13}{3}x + 7 \\ (-2y)/-2 = (\frac{13}{3}x + 7)/-2 \\ y = \frac{-13}{6}x - \frac{7}{2}

Plug in (-13/6)x - (7/2) for y in the second equation:

3x+ \frac{2}{5}y=5 \\ 3x+ \frac{2}{5}(\frac{-13}{6}x - \frac{7}{2})=5

Solve for x:

3x+\frac{2}{5} y=5 \\ [tex]3x+\frac{2}{5} (\frac{-13}{6}x - \frac{7}{2})=5 \\ 3x + \frac{-26}{30}x- \frac{14}{10} =5 \\ \frac{90}{30}x +\frac{-26}{30}x- \frac{42}{30} = \frac{150}{30} \\ \frac{64}{30}x - \frac{42}{30} = \frac{150}{30} \\ (\frac{64}{30}x - \frac{42}{30})+\frac{42}{30} = (\frac{150}{30})+\frac{42}{30} \\ \frac{64}{30}x= \frac{192}{30} \\ (\frac{64}{30}x)*30= (\frac{192}{30})*30 \\ 64x=192 \\ 64x/64 = 192/64 \\ x=3

Plug 3 in for x in the first equation:

\frac{-13}{3}x-2y= 7 \\ \frac{-13}{3}*3-2y= 7

Solve for y:

\frac{-13}{3}*3-2y= 7 \\ -13-2y=7 \\ (-13-2y)+13=(7)+13 \\ -2y=20 \\ (-2y)/-2=(20)/-2 \\ y = -10

For this system of equations, x = 3 and y = -10.

Hope this helps!

complete the steps to find the area of the trapezoid.

area of rectangle = 256   square units

area of triangle 1 = 24  square units

area of triangle 2 = 16  square units

area of triangle 3 =   96 square units

area of trapezoid = 120  square units

step-by-step explanation: