Solving Equations.Eliminaion method?

Substitue 4y-2 or 6y+8 into one of the equations, whichever one you choose will be fine because it will be the same equation no matter what, just backwards. I'll try to simplify it more: eq. 1) x=4y-2 eq. 2) x=6y+8 Substitute either 4y-2 or 6y+8 into the other equation. If you use 4y-2 from eq. 1, eq. 2 would now look like this: 4y-2=6y+8 because if x is 4y-2, you can just put it right into where the x is in eq. 2 - because x IS 4y-2. It is the same for if you use 6y+8 from eq. 2 -- x IS 6y+8 so you can put this phrase into equation one where the x is because 6y+8 equals x. It would look like this - 6y+8=4y-2 Then after you have either 6y+8=4y-2 or 4y-2=6y+8, you can solve for y by adding together like terms. Using 4y-2=6y+8...... Add the 4y and 6y together by subtracting 4y from the positive 4y on the left side of the equal sign and by 6y to get 2y on the right side of the equal sign : 4y-2=6y+8 -4y -4y ^ the +4y and -4y cancels out leaving... -2=2y+8 you are trying to get y by itself, so you need to subtract 8 from the positive 8 on the right side and subtract 8 from the negative 2 on the left to get -10 on the left: -2=2y+8 -8 -8 ^ the +8 and - 8 cancels out leaving.... -10=2y To get y by itself, you need to divide the 2y by 2 (do the opposite to cancel the 2 out, like we did when it was addition/subtraction), and divide the -10 by 2: -10=2y -10/2=2y/2 ^the multiplied 2 and divided 2 cancels out leaving.... -5=y <--- -5 IS y ---- So now that you have y, you can put in -5 in either eq.1 or 2 and solve : (using eq 1): x = 4y-2 x = 4(-5)-2 x = -20-2 x = -22 ! So the ordered pair to solve these equations is (x, y) : (-22, -5)

There are several things you can do but the easiest is to first clear each equation of fractions.

Multiply the first by 60, the LCM of 5,4,3 giving 12x - 15y = 20

Multiply the second by 6 giving 3x + 2y = 5 and multiply this by 4 to get 12x,

12x+8y=20. Now subtract this equation from that on line 2,

to give -23y=0 so y=0 then sub back giving x=5/3.

If you are nifty with fractions you could multiply equ 1 by 4: (4/5)x - y=4/3

and equ 2 by 3: (3/2)x+y=5/2 then add the equations

(4/5+3/2)x =(4/3+5/2) so (23/10)x=23/6 giving x = (23/6)/(23/10)=10/6=5/3 etc.

Multiply the first equation be 5 and the last equation by 2 to eliminate the x denominators:

x + 5y/4 = 5/3

x + 2y/3 = 5/3 => 10/6 = 5/3

Take the equations to eliminate the x variable:

0 + 5y/4 - 2y/3 = 0

Now, multiply by 4 then by 3 to cancel the denominators:

3(5y) - 4(2y) = 0

15y - 8y = 0

7y = 0

y = 0

Subbing back y = 0 in the latter provides:

x/2 = 5/6

x = 5/3

(x, y) = (5/3, 0)

You could of graphed these equations and find the intersecting point ;)

What we have basically done here is manipulate the WHOLE equation to single out one variable in order to cancel it when subtraction has occurred. However, we could have canceled out the y's. But, if you are beginning to learn this, I recommend that you multiply each equation be its LCM.

For instance, the latter implies that the lcm of 2 and 3 is 6.

So, you multiply the whole by 6:

3x + 2y = 5

The lcm of the former is 20, so multiply by 20 yields:

4x - 5y = 20/3

We can now multiply be 3 also:

12x - 15y = 20.

Now our equations are:

3x + 2y = 5

12x - 15y = 20

Notice that if we multiply the first be 4, then we have:

12x + 8y = 20

12x - 15y = 20

We can cancel the x's leaving

8y -(-15y) = 0

23y = 0

y = 0

Subbing back to either equations:

12x = 20 => x = 20/12=5/3

By the way, the LCM stands for Lowest common multiple which means the lowest number that divides all numbers in the set.

i