solving simultaneous equations?

Question

can ayone solve simultaneous equations showing working:
1.3x+5y=21
2x-3y=13

2. 2x-3y=13
5x+4y=18.5

3. 5x+y=16
x-2y=1

4. x-y=2
xy=15

let me see how good you guys really are as my friend claim

Answer 1

4.
x - y = 2
xy = 15

Solve the first equation in terms of x, and then plug in this value into the second equation.

x - y = 2
x = 2 + y

xy = 15
(2 + y)y = 15
2y + y^2 = 15
y^2 + 2y - 15 = 0
(y - 3)(y + 5) = 0, therefore
y = {3, -5}

When y = 3, x = 2 + y = 2 + 3 = 5
When y = -5, x = 2 + y = 2 + (-5) = -3

Therefore, you have two sets of solutions:
x = 5, y = 3
x = -3, y = -5

Answer 2

3x + 5y = 21
2x - 3y = 13

6x + 10y = 42 (1)
6x - 9y = 39 (2)

(1) - (2)
10y + 9y = 42 - 39
19y = 3
y = 3/19

2x - 3y = 13
2x - 3(3/19) = 13
2x - 9/19 = 13
2x = 256/19
x = 128/19 = 6 14/19

x = 6 14/19, y = 3/19
-----------------------------
2x - 3y = 13
5x + 4y = 37/2

10x - 15y = 65 (1)
10x + 8y = 37 (2)

(2) - (1)
8y + 15y = 37 - 65
23y = -28
y = -28/23 = -1 5/23

2x - 3y = 13
2x - 3(-28/23) = 13
2x + 84/23 = 13
2x = 215/23
x = 215/46 = 4 31/46

x = 4 31/46, y = -1 5/23
-----------------------------
5x + y = 16 (1)
x - 2y = 1

x = 2y + 1 (2)

Substitute (2) into (1)

5(2y + 1) + y = 16
10y + 5 + y = 16
11y = 11
y = 1

x - 2y = 1
x - 2 = 1
x = 3

x = 3, y = 1
-----------------------------
x - y = 2
xy = 15 (1)

x = y + 2 (2)

Substitute (2) into (1)

(y + 2)y = 15
y² + 2y = 15
y² + 2y - 15 = 0
(y - 5)(y + 3) = 0
y = 5 or y = -3

xy = 15
5x = 15
x = 3
or
-3x = 15
x = -5

x = 3, y = 5
x = -5, y = -3

Answer 3

3. 5x + y = 16 and x - 2y = 1 > (Solve for "x" or "y" in either equation > let's solve for "x" in the 2nd equation).

First: add 2y to both sides >

x - 2y + 2y = 1 + 2y
x = 1 + 2y

Sec: replace 1 + 2y with the "x" variable in the 1st equation >

5(1 + 2y) + y = 16
5(1) + 5(2y) + y = 16
5 + 10y + y = 16

*Combine "like" terms >

5 + 11y = 16

*Subtract 5 from both sides >

5 - 5 + 11y = 16 - 5
11y = 11

*Solve for "y" by isolating it on one side > divide both sides by 11

11y/11 = 11/11
y = 1

Third: replace 1 with the "y" variable in the 2nd equation >

x - 2(1) = 1
x - 2 = 1

*Add 2 to both sides >

x - 2 + 2 = 1 + 2
x = 3

(3, 1)

Follow the same format for the rest-give them a try :-)

Answer 4

1)3x+5y=21
2x-3y=13
Multiply the first equation by 3 and the second equation by 5:
9x+15y=63
10x-15y=65
=========
19x=128
x=128/19=6 14/19

2(128/19)-3y=13
256/19-3y=13
256/19-3y=247/19
-3y=-9/19
y=9/57=3/19

The solution set is (128/19,3/19)

Check:
3(128/19)+5(3/19)=21
384/19+15/19=21
399/19=21
21=21

2(128/19)-3(3/19)=13
256/19-9/19=13
247/19=13
13=13

2)2x-3y=13
5x+4y=18.5
Multiply the first equation by 4 and the second equation by 3:
8x-12y=52
15x+12y=55.5
===========
23x=107.5
x=107.5/23=4 15.5/23

2(215/46)-3y=13
430/46-3y=598/46
-3y=168/46
y=-56/46=-28/23=-1 5/23

The solution set is (107.5/23,-28/23)

Check:
2(107.5/23)-3(-28/23)=13
215/23+84/23=13
299/23=13
13=13

5(107.5/23)+4(-28/23)=18.5
537.5/23-112/23=18.5
425.5/23=18.5
18.5=18.5

3)5x+y=16
x-2y=1
Multiply the first equation by 2:
10x+2y=32
x-2y=1
========
11x=33
x=3

3-2y=1
-2y=-2
y=1

The solution set is (3,1)

Check:
5(3)+1=16
16=16

3-2(1)=1
3-2=1
1=1

4)x-y=2
xy=15

Add y from both sides:
x=y+2

Now, substitute it in the first equation:
(y+2)y=15
y^2+2y-15=0
(y+5)(y-3)=0
y=-5 and 3

-5x=15
x=-3

3x=15
x=5
x=-3 and 5

The solution sets are (-3,-5) and (5,3)

Check:
-3-(-5)=2
-3+5=2
2=2

(-3)(-5)=15
15=15

5-3=2
2=2

(3)(5)=15
15=15

I hope this helps!