Download 1078 Sec1 Elementary..

Solve for system of linear equations;
1.x+y=5 (1)
x-y=1 (2)  y=x-1 , insert the expression of y into (1)
 x+ (x-1) = 5  x=3 and y=2
4. 3x-2y=12 (1)
+ 7x+2y=8 (2)
10x =20  x=2
insert the value of x into (1)  3*2 – 2y =12 y=-3
So x=2 and y = -3
6. 3u + 5v = 15
(1)
6u + 10v = -30 (2)
(1)*2 so that make the coefficient in front of v the same for both equations.
 6u+10v=30 (1’)
(-) 6u +10v=-30 (2)
0= 60  can not be realized, there is no solution.
Actually, we can find there is no solution at the early beginning. If draw the equation as
the linear line in the coordinate system, we can find that they have the same slope and
different intercept  they are two parallel lines.
8. y=x-4 (1)
x+3y=12 (2)  insert the expression of y into (2)
 x+ 3(x-4)=12  4x=24  x=6
y = x-4 = 2
so x=6 and y =2
10. 3x-y=7 (1)  y=3x-7
2x+3y=1 (2)
insert y=3x-7 into the (2) equation  2x + 3(3x-7) =1  11x=22  x=2
y=3x-7= 3*2-7=-1
so x=2 and y=-1
12. 2x-3y=-8 (1)
(+) 5x+3y =1 (2)
7x= -7 x=-1 as 2x-3y=-8  -2-3y=-8 y=2
so x=-1 and y=2
1
16. 4x+3y=26 (1)
3x-11y= -7 (2)
(1)*3  12x+9y=78
(-) (2)*4  12x-44y=-28
53y = 106 y=2
4x+3y=26  4x+6 = 26  x=5
so x=5 and y=2
18. 3x-6y= -9 (1)
-2x+4y=12 (2)
(1)/3  x-2y=-3
(2)/-2  x-2y= -6
These two lines are parallel to each other. So there is no solution.
From the original form it is difficult to see the relation, so I should normalize the
coefficient in front of either x or y.
24. y=0.08x
(1)
y=100 +0.04x (2)
Equalize (1) and (2).
0.08x=100+0.04x  0.04x=100  x=2500
7
5
x  y  10
2
6
30.
2
4
x y 6
5
3
(1)
(2)
Solution 1:
2
2 5
2
5
20
(1)*
 x   y  10  x 
y
7
7 6
7
21
7
5
5 4
5
10
(2)*
 x   y   6  x   y  15
2
2 3
2
3
5
20
10
5 70
85
75
85
85 21 17

y
  y  15  (  ) y 

y
y * 
21
7
3
21 21
7
21
7
7 75 5
Insert back to solve for x: x 
5
20
17
y
and y 
21
7
5
 x= 11/3
2
Solution 2:
(1) *6  21x  5 y  60
(1')
(2)*15  6 x  20 y  90 (2 ')
 (1') * 4  84 x  20 y  240 plus with (2 ')
90 x  330  x 
y
11
3
17
5
6
36. (1) 2x+3y=18  y=-2/3X+6
(2) 2x-6y= -6  y=1/3x+1
(3) 4x+6y= -24  y=-2/3x-4
1
-6
-3
9
Line (1) meets (2)
Line (2) meets line (3)
-4
As long as (1) and (3) are parallel  they cannot meet each other. There is no solution
between (1) and (3).
(1) and (2) meet at when x=5 and y=8/3
(2) and (3) meet at when x=-5 and y=-2/3
38. p=0.4q+3.2 supply
p=-1.9q+17 demand
Set supply = demand  0.4q+3.2=-1.9q+17  q=6 and p=5.6
3