5x+4y=2 and 2x-6y=-3..?

Hi,

5x+4y=2 and 2x-6y=-3

Solve this by substitution by first solving for a variable in either equation. then substitute that expression into the other equation and solve.

Solve 2x-6y= -3 for x.
2x - 6y = -3
2x = 6y - 3
x = 3y - 3/2

Substitute 3y - 3/2 into the first equation in place of x.
5x+4y=2
5(3y - 3/2) + 4y = 2
15y - 15/2 + 4y = 2
Multiply every term by 2 to eliminate the fractions.
30y - 15 + 8y = 4
38y - 15 = 4
38y = 19
y = 19/38
y = ½

If y = ½ and 5x + 4y = 2, then:
5x + 4(½) = 2
5x + 2 = 2
5x = 0
x = 0

(0,½) <==ANSWER

I hope that helps!! :-)

Relative values of y:
5x + 4y = 2
4y = 2 - 5x
y = (2 - 5x)/4

2x - 6y = - 3
6y = 2x + 3
y = (2x + 3)/6

Value of x:
6(2 - 5x) = 4(2x + 3)
3(2 - 5x) = 2(2x + 3)
6 - 15x = 4x + 6
19x = 0
x = 0

Value of y:
= (2 - 5[0])/4
= (2 - 0)/4
= 2/4
= 1/2

Answer: x = 0, y = 1/2

Proof (1st equation):
5(0) + 4(1/2) = 2
0 + 2 = 2
2 = 2

Proof (2nd equation):
2(0) - 6(1/2) = - 3
0 - 3 = - 3
- 3 = - 3

5x + 4y = 2____(1)
2x - 6y = -3____(2)
from equation (1), x = 2/5 - 4y/5___(3)
substitute equation (3) into equation (2);
2(2/5 - 4y/5) - 6y = -3
[2(2 - 4y)]/5 = 6y - 3
4 - 8y = 30y - 15
4 + 15 = 30y + 8y
38y = 19
y = 1/2
substitute y = 1/2 into equation (3);
x = 2/5 - [4(1/2)]/5
x = 2/5 - 2/5
x = 0

5x+4y=2 > 5x = 2-4y > x = 2/5 - 4/5y
2x-6y=-3

2(2/5-4/5y)-6y=-3
4/5-8/5y-6y=-3
-38/5y=-3 4/5 = -19/5
y = 19/5(5/38) = 19/38 = 1/2

5x + 4(1/2) = 2
5x + 2 = 2
x = 0

I'll help, but that's it. Because it's YOUR assignment.

1: 5x + 4y = 2
1': 5x - 2 = -4y
4y = -5x + 2
y = -5x/4 + 2/4
= -5x/4 + 1/2

2: 2x - 6y = -3

Then sub 1' into 2.

5x+4y=2 and 2x-6y=-3
5x+4y=2
=> x = 2 - 4y/5

2x-6y=-3

2(2 - 4y/5) -6y = -3

4 - 8y
-------- -6y = -3
5

4 - 8y -30y = -3
------------
5

4 - 38y = -15
-38 y = -19
y = -19 / -38

y = 1/2

5x + 2 = 2

5x = 0

x = 0



Hope this helps

If you start with the second equation and isolate x, then you have
x = 3y -3/2
Subbing this into the first one gives:
15y -15/2 + 4y = 2
so 19y = 9.5
>> y = 0.5
Sub this value for y back into either equation, say the first one:
>>5x + 2 = 2
>> x = 0

15x + 12y = 6
4x - 12y = - 6--------Add

19x = 0
x = 0

12y = 6
y = 1/2

x = 0 , y = 1/2

5x + 4y = 2
2x - 6y = -3

2x - 6y = -3
2x = -3 + 6y
x = (6y - 3)/2
5x = 5(6y - 3)/2

5x + 4y = 2
5(6y - 3)/2 + 4y = 2
(30y - 15)/2 + 4y = 2
2[(30y - 15)/2 + 4y] = 2(2)
30y - 15 + 8y = 4
30y + 8y = 4 + 15
38y = 19
y = 19/38
y = 1/2 (0.5)

2x - 6y = -3
2x - 6(1/2) = -3
2x - 3 = -3
2x = -3 + 3
2x = 0
x = 0/2
x = 0

∴ x = 0 , y = 1/2 (0.5)

first you need to make the two equations match a number. I suggest Y. Multiply all the numbers on the right equation by two. gives you 4x-12y=-6. Multiply all numbers on the left by three. 15x+12y=6. Notice how the y's are the same now? solve for y on either side. lets do the left side. 12y=6-15x, (6-15x)/12 and plug it in on the right. 2x-12(6-15x)/12=-6. The twelves cancel, (convenient this will happen all the time because you intentionally made those numbers match) leaves you with 2x-(6-15x)=-6 which is really 2x-1(6-15x). multiply your parentheses with -1 to make it positive and remove the parentheses 2x+6+15x=6 (remember one side of the equation must happen to the other or it is no longer true) leaves you with 6+17x=6, 17x=12 x=12/17.

You already solved for y remember? 6-15x= 6-15(12/17)=y (might want to simplify these before you turn it in.