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!

answer:

c. the third graph

54 kilometers per hour

step-by-step explanation:

according to rules:

1 km = 1000 meter

which means that 1 meter = 1/1000 km

1 hour = 3600 second [ as 1 hour = 60 min = 60 *60 seconds ]

which means that 1 second = 1/3600 hour

therefore, 1 kilometer/hour =

thus to convert 15 meter per second to kilometer per hour, we multiply 15 to 3.6 kilometer/hour.

15 meter/second=   which is the required answer.