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# Solve the given linear system.

Solve the given linear system.

a. y = 5x - 3
y = 3x - 1

b. -2x + 3y = 8
3x + 5y = -12

c. 2x - 4y = -6
-x + 2y = 3

d. 5x + 10y = 70
5x + 25z = 270
10y + 25z = 300

e. 2x - y - z = 5
x + y + z = 7
3x - 2y - 3z = 1

### 2 Answers by Expert Tutors

Priti S. | Math Tutor with Patience & KnowledgeMath Tutor with Patience & Knowledge
5.0 5.0 (452 lesson ratings) (452)
1
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

5x - 3 = 3x - 1
-3x       -3x
______________
2x - 3 = - 1
+3     +3
______________
2x = 2
x = 1
Plug in the value of x to find y
y = 5(1) - 3
y = 2

(1,2)
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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 )
_________________

-6x + 9y = 24
6x +10y = -24
_________________ now x can be eliminated.
19y = 0
y = 0
Plug in the value of y to find x
-2x + 3y = 8
-2x + 3(0) = 8
-2x  = 8
x = -4

(-4,0)
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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             +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.)

infinite many solution

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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

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

z = 10  plug in z to find y

10y + 25z = 300
10y + 25(10) = 300
10y + 250 = 300
-250     -250
10y = 50

y = 5 plug in y to find x

5x + 10(5) = 70

5x + 50 = 70
-50     -50
5x = 20
x = 4

(4, 5, 10)

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e.   Eliminate z & y from first 2 equation
2x - y - z = 5
x + y + z = 7
3x - 2y - 3z = 1

2x - y - z = 5
x + y + z = 7
__________

3x = 12
x = 4    Now take 2nd and 3rd equation and plug in x

x + y + z = 7  >>>>>>>>>>>   4 + y + z = 7   >>>>>>>     y + z = 3
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
_____________
-z = -5

z = 5  Now plug in z and x value to find y

x + y + z = 7
4 + y + 5 = 7
9 + y = 7
-9         -9
y = -2

(4, -2, 5)

Michael F. | Mathematics TutorMathematics Tutor
4.7 4.7 (7 lesson ratings) (7)
0
y = 5x - 3
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
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
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