Urgent math help (Equation related)

[KillerJuan77]

Ok, so I need to solve this 3 equations with the sustitution method:

5x + 8y = 136
3x + 10y = 118

x + y = 40
2x + 4y = 132

a + n = 224
35a + 15n = 4480

And the big problem is that I don't remember how to solve them, can someone help me?

[Ivo]

Ok, so I need to solve this 3 equations with the sustitution method:

5x + 8y = 136
3x + 10y = 118

x + y = 40
2x + 4y = 132

a + n = 224
35a + 15n = 4480

And the big problem is that I don't remember how to solve them, can someone help me?

It is pretty simple. The point is that you solve one of the equations for one of the variables and then put it back in the other one to make the 2nd one an equation with just one variable. I'll show you explicitly for one of them but you will have to do the rest:

x + y = 40
2x + 4y = 132

First step can be
x=40-y

Second step is put that into the other one, where x is:

80 - 2y + 4y =132

which means you know can know what y is:

2y = 132 -80 => y=26

The final step is going back to the first step now that you know y, you know x
x=40-26 => x=14.

Ivo.

[flojocabron]

too much math makes my head hurt!

[Ivo]

too much math makes my head hurt!

Dude, this is pretty basic math... And helpful in real life as well.

Ivo.

[KillerJuan77]

It is pretty simple. The point is that you solve one of the equations for one of the variables and then put it back in the other one to make the 2nd one an equation with just one variable. I'll show you explicitly for one of them but you will have to do the rest:

x + y = 40
2x + 4y = 132

First step can be
x=40-y

Second step is put that into the other one, where x is:

80 - 2y + 4y =132

which means you know can know what y is:

2y = 132 -80 => y=26

The final step is going back to the first step now that you know y, you know x
x=40-26 => x=14.

Ivo.

Sorry to ask but where did the 80 come from?

[YoshiEgg25]

It is pretty simple. The point is that you solve one of the equations for one of the variables and then put it back in the other one to make the 2nd one an equation with just one variable. I'll show you explicitly for one of them but you will have to do the rest:

x + y = 40
2x + 4y = 132

First step can be
x=40-y

Second step is put that into the other one, where x is:

80 - 2y + 4y =132

which means you know can know what y is:

2y = 132 -80 => y=26

The final step is going back to the first step now that you know y, you know x
x=40-26 => x=14.

Ivo.

Sorry to ask but where did the 80 come from?

If x = 40 - y
then 2x = 2 (40-y)
which means x = 80 - 2y

[Hatta]

Substitution is an incredibly powerful technique. Most of math is just rearranging statements and plugging them into each other.

[gtmtnbiker]

Maybe this might make it a little clearer.

x + y = 40
2x + 4y = 132

The idea is that you need to solve for one variable. Whatever you do needs to be applied to one side.

So with:
x + y = 40

we want x or y to be alone. So we'll subtract y from both sides.

x + y - y = 40 - y
x + (y - y = 0) = 40 - y
x = 40 - y

Now we know that x is equal to 40 - y.

The second equation is:
2x + 4y = 132

We know x = 40 -y. So wherever you see x in the second equation, you substitute 40 -y.

So
2x + 4y = 132
is
2 (40 - y) + 4y = 132
is
2 x 40 - 2 x y + 4y = 132
80 - 2y + 4y = 132
80 + 2y = 132

now we can subtract 80 from both sides
80 - 80 + 2y = 132 - 80
0 + 2y = 52
2y = 52
now we want to divide both sides by 2 to find the value of y
2y/2 = 52/2
y = 26

Now we know y is 26. We can plug it into either equation to find out x.

x + y = 40
x + 26 = 40
subtract 26 from both sides
x + 26 - 26 = 40 - 26
x + 0 = 14
x = 14

[Bradtemple87]

Hmm, well let me know when you need help with your English essay because when it comes to math I am long gone.

[MrPopo]

Another way to view it is that you're just using Gaussian elimination.

5x + 8y = 136 (L1)
3x + 10y = 118 (L2)

The goal is to eliminate x from the bottom equation, then y from the top equation (this generalizes for a system of n equations with n unknowns to only have the 1st term in the 1st equation, only the 2nd term in the 2nd equation, etc).

So you add -3/5 * L1 to L2. This will eliminate the x term, and it looks like this:

(-3x - 12/5y = -408/5) = L1, this is added to L2, so L2 becomes (38/5y = 182/5)

Now you eliminate the y term from L1. You multiply the new L2 by -40/38 so that it will eliminate y from the first term. So you get:

(-8y = -728/19) = L2, which is added to the original L1, so L1 becomes (5x = 1856/19).

So now your final system looks like:

5x + 0y = 1856/19
0x + 38/5y = 182/5

Solve for x and y.