solve 7x/3 + 5y?4 = 4, 5x/6 -2y = 21 by substitution

[Helen]

I posted a new message earlier thanking Eliz. Stapel and Subhotosh Khan for trying to help me with my algebra problem. I don't know what happened to it but I can't find it anywhere. I typed out the whole problem again but can't find it. I will do it again.
*Clear fractions, solve by substitution.
(1) 7x/3 + 5y?4 = 4 (2) 5x/6 -2y = 21
*multiply by 12.
(3) 28x + 15y = 48
*solve equation (3) for x.
28x = -15y + 48
15y + 48
----------------------------
28 = 28
*substitute 15y + 48 forx in (4)
(4) 5x -12y = 126
5(15y+48) -12 =126
5-5 -12 + 48 = 126
-3 + 48 = 126
45y = 126
5y = 14
y = -14
*replace y with -14 in either . (3) or (4).
28x + 15y = 48
28x + 15(-14) = 48
28X + 1 = 48
28X +1-1 = 48
28X = 48
7X = 12
X = -12

 

[stapel]

I posted a new message earlier....I don't know what happened to it but I can't find it anywhere.

Your earlier thread was, when you started this thread, the fourth message from the top. I don't know why it is not visible in your browser...? In future, you might try going to your "profile" and clicking through to your messages directly, thus avoiding whatever error your browser is experiencing. :idea:

*Clear fractions, solve by substitution.
(1) 7x/3 + 5y?4 = 4 (2) 5x/6 -2y = 21

Are these two exercises, Exercise (1) and Exercise (2), or are you numbering two equations in one linear system? Is the question-mark in (1) meant to be a division "slash"?

*multiply by 12.

Is this part of the instructions, a step you took, or something else? What is being multiplied by 12? Equation (1)?

As mentioned in your other thread:

...your multi-line forced-space formatting does not display clearly. You might want to try using the formatting explained at the links provided in the "Read Before Posting" thread....

I think you mean the following:

. . . . .Solve the system of equations, after first clearing
. . . . .the fractions, if any.

. . . . .(7/3)x + (5/4)y = 4
. . . . .(5/6)x - 2y = 21

. . . . .Multiplying the first equation by 12, I get:

. . . . .28x + 15y = 48
. . . . .(5/6)x - 2y = 21

I'm not sure what followed in your post, but the next step, according to the instructions, would be:

. . . . .Multiplying the second equation by 6, I get:

. . . . .28x + 15y = 48
. . . . .5x - 12y = 126

Using substitution (solving one equation for one variable, and plugging that into the other equation) will clearly quickly get messy. A better method might be elimination/addition:

. . . . .Multiply the first equation by 4 and the second by 5:

. . . . .112x + 60y = 192
. . . . .25x - 60y = 630

Then add the equations. Solve the result for the value of x. Back-solve for the value of y. :wink:

Eliz.

 

[Helen]

Intermediate Algebra

Eliz. Stapel, Just a late thank you. Helen