solve 7x/3 + 5y/4 = 4, 5y/6 - 12y = 21 by substitution

[Helen]

Clear fractions, solve by substitution.
(1) (2)
7x 5y 5y * multiply by 12
---- + ---- = 4 ---- 12y = 21
3 4 6

(3) 28x + 15y = 48 (4) 5x - 12y = 126
5 (15y+48) -12y = 126
*solve equation (3) for x. 5-5-12 +48 = 126
-3 + 48 = 126
28x = -15y + 48 45y = 126
15y + 48 5y = 14
------------------------ y = -14
28x = 28

* substitute 15y + 48 for x in (4)

*replace y with -14 in either equation, (3) or (4).


28x + 15y = 48
28x + 15(-14) = 48
28x + 1-1 = 48
28x = 48
7x = 12
x = -12

Can you tell me if I am doing this in the right way? Right answer?

 

[stapel]

As you noticed when you reviewed your post, 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 you read.

Clear fractions, solve by substitution.
(1) (2)
7x 5y 5y * multiply by 12
---- + ---- = 4 ---- 12y = 21
3 4 6

I think the above means the following:

. . . . .1) Solve (7x)/3 + (5y)/4 = 4

. . . . .2) Solve [(5y)/6)][12y] = 21

Note: The first equation has two variables, and thus cannot be solved for a numercial answer.

(3) 28x + 15y = 48 (4) 5x - 12y = 126
5 (15y+48) -12y = 126
*solve equation (3) for x. 5-5-12 +48 = 126
-3 + 48 = 126
28x = -15y + 48 45y = 126
15y + 48 5y = 14
------------------------ y = -14
28x = 28

I think the above means the following:

. . . . .3) Solve 28y + 15y = 48

. . . . .4) Solve 5x - 12y = 126

Note: These have the same difficulty as does exercise (1) above. And I'm afraid I can't decipher what follows the equations.

Please reply with clarification. Thank you!

Eliz.

 

[Deleted member 4993]

Re: Intermediate Algebra

Clear fractions, solve by substitution.
(1) (2)
7x 5y 5y * multiply by 12
---- + ---- = 4 ---- 12y = 21
3 4 6
_________________________________________________
Does your problem look like:

\(\displaystyle \frac{7x}{3} + \frac{5y}{4} = 4\)

and

\(\displaystyle \frac{5x - 12y}{6} = 21\)

___________________________________________________
(3) 28x + 15y = 48 (4) 5x - 12y = 126
5 (15y+48) -12y = 126
*solve equation (3) for x. 5-5-12 +48 = 126
-3 + 48 = 126
28x = -15y + 48 45y = 126
15y + 48 5y = 14
------------------------ y = -14
28x = 28

* substitute 15y + 48 for x in (4)

*replace y with -14 in either equation, (3) or (4).


28x + 15y = 48
28x + 15(-14) = 48
28x + 1-1 = 48
28x = 48
7x = 12
x = -12

Can you tell me if I am doing this in the right way? Right answer?

 

[Helen]

Intermediate Algebra

Subhotosh Khan, Just a late thank you. Helen