systems of equations - substitution

[Guest]

I am kind of new to " Free Math Help " and after reading through everything, I might still be a little confused, so please don't get mad at me ? I need help with this summer packet. The page says : " SOLVE EACH SYSTEM OF EQUATIONS BY THE SUBSTITUION METHOD: example : (1) y=5-2x (2) 5x-6y=21 " I am fine with the easy ones that tell you what's y and what's x but I'm not sure about a question like " (1) 8x+3y=2 (2) 2x=y-4. I tried working it out like the example but my answers are WAY OUT THERE! So if anyone can help explain it to me, that'd be great!

Thanks,
ann.banana

 

[jonboy]

LOL I really looooove the name. You got my respect.


\(\displaystyle \L (1)\,8x\,+\,3y\,=\,2\)


\(\displaystyle \L (2)\,2x\,=\,y\,-\,4\)


Ok since the x is very close from getting islolated divide equation (2) by 2:\(\displaystyle \L \;x\,=\,\frac{y\,-\,4}{2}\)

Subsitute that back in equation (1): \(\displaystyle \L \;8(\frac{y-4}{2})\,+\,3y\,=\,2\)


Solve for y and then back solve for x. If you get stuck show use what appears to be a problem. Also plug your variables back in to make sure everything is correct.

 

[jonboy]

hey; thanks for the help.
even though you gave me some great advice...i'm still a little confused.
i'm not that great at math and i was wondering if you can give me a few extra steps? i don't want you to do it for me, but you get my point ?

thanks so much,
ann<3banana

Ok the new equation can be written as \(\displaystyle \L \;\frac{8}{1}(\frac{y-4}{2})\,+\,3y\,=\,2\)

What is: \(\displaystyle \L \;(\frac{8}{1})(\frac{y-4}{2})\,+\,3y\,=\,2\)

Then: Solve for y and then back solve for x. If you get stuck show use what appears to be a problem. Also plug your variables back in to make sure everything is correct.

No more steps till you show use what you have done.

 

[jonboy]

So now,
what i got was:

8 over 1 x y-4 over 2 =
4y-32+3y=26

You add like terms. [ 7y and 58 ]
And when you divide 58 by 7 it doesn't come out perfect. what do you do in a situation like that?

Sry you are wrong.

Ok for (8/1) * (y-4/2) = 8(y-4)/2 = ?

Also like terms are like 4x and 3x, 2 and 1, 2x^2 and 3x^2.

 

[Denis]

I'm not sure about a question like " (1) 8x+3y=2 (2) 2x=y-4. I tried working it out like the example but my answers are WAY OUT THERE!

...why is this thread so confusing; Jonboy, where d'ya get them quotes?

ann, your 2 equations are:
8x + 3y = 2 [1]
2x - y = -4 [2]
Multiply [2] by 3:
6x - 3y = -12 [2] ; bring in [1]:
8x + 3y = 2 [1]
Now, add them up:
14x = -10
x = -10/14
x = -5/7 : OK :?: :?:

 

[jonboy]

I'm not sure about a question like " (1) 8x+3y=2 (2) 2x=y-4. I tried working it out like the example but my answers are WAY OUT THERE!

...why is this thread so confusing; Jonboy, where d'ya get them quotes?

ann, your 2 equations are:
8x + 3y = 2 [1]
2x - y = -4 [2]
Multiply [2] by 3:
6x - 3y = -12 [2] ; bring in [1]:
8x + 3y = 2 [1]
Now, add them up:
14x = -10
x = -10/14
x = -5/7 : OK :?: :?:

I got the quotes because Ann kept on PMing me to look what she had done instead of just posting it here. Also I would have done what you'd done but that is not subsitution.

 

[Mrspi]

LOL I really looooove the name. You got my respect.


\(\displaystyle \L (1)\,8x\,+\,3y\,=\,2\)


\(\displaystyle \L (2)\,2x\,=\,y\,-\,4\)


Ok since the x is very close from getting islolated divide equation (2) by 2:\(\displaystyle \L \;x\,=\,\frac{y\,-\,4}{2}\)

Subsitute that back in equation (1): \(\displaystyle \L \;8(\frac{y-4}{2})\,+\,3y\,=\,2\)


Solve for y and then back solve for x. If you get stuck show use what appears to be a problem. Also plug your variables back in to make sure everything is correct.

What you've done is correct, jonboy....but I think there is an easier approach. If you are trying to solve a system by substitution, spend a moment looking at the equations to see if one of them can easily be solved for one of the variables without introducing fractions. An equation with no fractions is easier for most folks to deal with.

You have

2x = y - 4

Do you see that this can be solved for y simply by adding 4 to both sides?

2x + 4 = y

Now, substitute (2x + 4) for y in the first equation:

8x + 3y = 2
8x + 3(2x + 4) = 2

Now, solve for x. Once you have the value for x, you can find y by using the equation
y = 2x + 4

 

[jonboy]

What you've done is correct, jonboy....but I think there is an easier approach. If you are trying to solve a system by substitution, spend a moment looking at the equations to see if one of them can easily be solved for one of the variables without introducing fractions. An equation with no fractions is easier for most folks to deal with.

You have

2x = y - 4

Do you see that this can be solved for y simply by adding 4 to both sides?

2x + 4 = y

Now, substitute (2x + 4) for y in the first equation:

8x + 3y = 2
8x + 3(2x + 4) = 2

Now, solve for x. Once you have the value for x, you can find y by using the equation
y = 2x + 4

Yes that would be more appropriate and less confusing. I just saw that x almost islolated and went straight for it! Good to have you here setting me straight.