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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
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2 Answers

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)
 
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

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