How do you solve the following linear system: 3x + 5y = -1, 2x − 5y = 16?

3 Answers
Apr 6, 2018

x = 3, y = -2

Explanation:

Starting with the first equation, we need to make either x or y the subject (I chose x)

3x + 5y = -1

x = (-5y-1)/3

2x - 5y = 16

2*(-5y-1)/3 - 5y = 16

(-10y-2)/3 - 5y = 16

-10y-2 - 15y = 48

-25y = 50

y = -2

Now we simply substitute (y = -2) into the second equation...

2x-5y = 16

2x-5(-2)=16

2x + 10 = 16

2x = 6

x = 3

Finally, we should also verify both answers by substituting both of them into the other equation again.

3x + 5y = -1

#3(3) + 5(-2) = -1

9 - 10 = -1

-1=-1, so therefore our answers are correct!

Apr 6, 2018

See a solution process below:

Explanation:

Step 1) First, solve each equation for 5y:

  • Equation 1:

3x + 5y = -1

3x - color(red)(3x) + 5y = -1 - color(red)(3x)

0 + 5y = -1 - 3x

5y = -1 - 3x

  • Equation 2:

2x - 5y = 16

2x - 5y + color(red)(5y) - color(blue)(16) = 16 - color(blue)(16) + color(red)(5y)

2x - 0 - color(blue)(16) = 0 + 5y

2x - 16 = 5y

5y = 2x - 16

Step 2) Because both equations are equal on the left side we can now equate the right sides of the equation and solve for x:

-1 - 3x = 2x - 16

-1 + color(blue)(16) - 3x + color(red)(3x) = 2x + color(red)(3x) - 16 + color(blue)(16)

15 - 0 = (2 + color(red)(3))x - 0

15 = 5x

15/color(red)(5) = (5x)/color(red)(5)

3 = (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))

3 = x

x = 3

Step 3) Substitute 3 for x in either of the solved equations in Step 1 and solve for y:

5y = 2x - 16 becomes:

5y = (2 xx 3) - 16

5y = 6 - 16

5y = -10

(5y)/color(red)(5) = -10/color(red)(5)

(color(red)(cancel(color(black)(5)))y)/cancel(color(red)(5)) = -2

y = -2

The Solution Is:

x = 3 and y = -2

Or

(3, -2)

Apr 6, 2018

x = 3 and y =-2

Explanation:

Notice that the terms in y in the two equations are are additive inverses. The sum of additive inverses is 0
Add the equations together to eliminate the y terms.

color(white)(xxxxxxxx)3xcolor(blue)(+5y) = -1" " A
color(white)(xxxxxxxx)2xcolor(blue)(-5y) = 16" "B

A+B:color(white)(xxx)5xcolor(blue)(+0y) = 15
color(white)(xxxxxxxxx)xcolor(white)(xxx) = 3

Substitute in A:
color(white)(xxxxxxxx)3(3)+5y = -1" " A
color(white)(xxxxxxxxxx)9+5y = -1
color(white)(xxxxxxxxxxxxx)5y = -10
color(white)(xxxxxxxxxxxxxx)y = -2

Check in B:
color(white)(xxxxx)2(3)-5(-2) = 16" "B
color(white)(xxxxxxxxxx)6+10 = 16
color(white)(xxxxxxxxxxxxx)16 = 16

The values are correct.