y = 3x - 1

3x + 5y = -12

-x + 2y = 3

5x + 25z = 270

10y + 25z = 300

x + y + z = 7

3x - 2y - 3z = 1

Solve the given linear system.

a. y = 5x - 3

y = 3x - 1

y = 3x - 1

b. -2x + 3y = 8

3x + 5y = -12

3x + 5y = -12

c. 2x - 4y = -6

-x + 2y = 3

-x + 2y = 3

d. 5x + 10y = 70

5x + 25z = 270

10y + 25z = 300

5x + 25z = 270

10y + 25z = 300

e. 2x - y - z = 5

x + y + z = 7

3x - 2y - 3z = 1

x + y + z = 7

3x - 2y - 3z = 1

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Marked as Best Answer

a. Use substitution to solve (substitute for y and set them equal to each other)

y = 5x - 3

y = 3x - 1

y = 3x - 1

5x - 3 = 3x - 1

-3x -3x

______________

2x - 3 = - 1

+3 +3

______________

2x = 2

Plug in the value of x to find y

y = 5(1) - 3

b. Use elimination to solve the problem (Try to get rid of one variable I picked x to get rid of)

(multiplying by 3 to the top Eq. & multiplying by 2 to the bottom Eq.)

3 ( -2x + 3y = 8 )

2 ( 3x + 5y = -12 )

2 ( 3x + 5y = -12 )

_________________

-6x + 9y = 24

6x +10y = -24

_________________ now x can be eliminated.

19y = 0

Plug in the value of y to find x

-2x + 3y = 8

-2x + 3(0) = 8

-2x = 8

------------------------------------------------------------------------------------------------

c. For this problem i will use substitution sove for x for the 2nd equation

2x - 4y = -6

-x + 2y = 3

-x + 2y = 3

-x + 2y = 3

+x +x

______________

2y = 3 + x

-3 -3

______________

x = 2y - 3 Now Plug in 2y -3 in the x for the 1st equation

2x - 4y = -6

2 (2y - 3) - 4y = -6

4y - 6 -4y = -6

+6 +6

0y = 0

0 = 0

when you have a 0 = 0 that means** infinite many solution** for this 2 equations ( coincided lines.)

---------------------------------------------------------------------------------------------------------------

d. I took 1st 2 equation and try to eliminate a variable x by multiplying by (-1) to 1st equation

5x + 10y = 70

5x + 25z = 270

10y + 25z = 300

5x + 25z = 270

10y + 25z = 300

-1 (5x + 10y = 70) >>>>>> -5x - 1oy = -70

5x + 25z = 270 >>>>>>>>> 5x + 25z = 270

5x + 25z = 270 >>>>>>>>> 5x + 25z = 270

-5x - 1oy = -70

5x + 25z = 270

_______________ now x is eliminated and new equation will be form

-10y + 25z = 200 ( now take the 3rd equation and do elimination y.

10y + 25z = 300

____________________

50z = 500

10y + 25z = 300

10y + 25(10) = 300

10y + 250 = 300

-250 -250

10y = 50

5x + 10(5) = 70

5x + 50 = 70

-50 -50

5x = 20

e. Eliminate z & y from first 2 equation

2x - y - z = 5

x + y + z = 7

3x - 2y - 3z = 1

x + y + z = 7

3x - 2y - 3z = 1

2x - y - z = 5

x + y + z = 7

x + y + z = 7

__________

3x = 12

x + y + z = 7 >>>>>>>>>>> 4 + y + z = 7 >>>>>>> y + z = 3

3x - 2y - 3z = 1 >>>>>>>>>> 3(4) - 2y - 3z = 1 >>>>> - 2y - 3z = -11

3x - 2y - 3z = 1 >>>>>>>>>> 3(4) - 2y - 3z = 1 >>>>> - 2y - 3z = -11

y + z = 3

- 2y - 3z = -11 Now eliminate y by mulitplying by 2 to the 1st equation

2( y + z = 3 )

- 2y - 3z = -11

2y + 2z = 6

- 2y - 3z = -11

- 2y - 3z = -11

_____________

-z = -5

x + y + z = 7

4 + y + 5 = 7

9 + y = 7

-9 -9

Michael F. | Mathematics TutorMathematics Tutor

y = 5x - 3

y = 3x - 1

y = 3x - 1

5x-3=3x-1

2x=2

x=1

y=5*1-3=2 or

y=3*1-1=2

-2x + 3y = 8

3x + 5y = -12

3x + 5y = -12

Multiply the first equation by 5 and the second by 3 to get

-10x+15y=40

9x+15y=-36

Subtract to get

-19x=76 or x=-4, which substituted in -2x + 3y = 8 gives

8+3y=8 or y=0

2x - 4y = -6

-x + 2y = 3

-x + 2y = 3

Multiply the second equation by -2 and get

2x-4y=-6, the first equation. The solution is any point on the line 2x - 4y = -6

or in standard form y=1/2x+3/2

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