Substitution Problem, I'm Stuck!

[KP]

I'm stuck on this problem: { y=2x
7x+5y=34

I'm not sure how to do this one, here's my work so far:
7x+5y=34
___ ___
5y 5y

7x=34-5y

Thats as far as I've gotten - help anyone? Thanks
KP

 

[stapel]

I'm stuck on this problem: { y=2x
7x+5y=34

Is this a system of equations? Does the above mean the following?

. . . . .Solve the system:
. . . . . . .y = 2x
. . . . . . .7x + 5y = 34

7x+5y=34
___ ___
5y 5y

Does the above indicate that you divided through on both sides by "5y"? If so, then the result would be:

. . . . .(7x)/(5y) + 1 = 34/(5y)

7x=34-5y

So you actually subtracted "5y" from either side...?

Since one of the equations is already solved for "y=", you might want to consider using the method of substitution, and plug the right-hand side of the first equation ("substitute") for "y" in the second equation. Solve for "x", and then back-solve for "y".

Eliz.

 

[KP]

I'm still having some trouble figuring this out

I have to figure what X and Y are in the problem,

was I doing it right ?

 

[stapel]

was I doing it right ?

The reason I requested clarification was that I couldn't tell. I apologize for the confusion.

Please follow the instructions in the previous reply. If you get stuck, please reply showing your steps. Thank you.

Eliz.

 

[KP]

i first did the 7x+5y=34

because five is the lowest i minused it from the left and right

with that i got the 7x =34-5y

after that im still stuck...

 

[stapel]

Since one of the equations is already solved for "y=", you might want to consider using the method of substitution, and plug the right-hand side of the first equation ("substitute") for "y" in the second equation. Solve for "x", and then back-solve for "y".

If you get stuck following these instructions, please reply showing your steps. Thank you.

Eliz.

 

[KP]

so do i substitute 2x for 5y???

 

[pka]

Use substitution: 7x+5(2x)=34.
Solve for x!
Use x to find y.

 

[KP]

how do i solve for X?

if I figure it out I can finish my homework

 

[Guest]

7x+5(2x)=34

7x +10x =34

17x =34

x=2

and so on....

 

[pka]

how do i solve for X? if I figure it out I can finish my homework

Oh my goodness, you have a great deal of work ahead of you.
You should not even try to solve a system of equations until you are completely comfortable with absolutely simple equations such as 7x+5(2x)=34.

If I were you, I would have a sit-down with my instructor.

 

[Guest]

y=2x -- equation (1)
7x+5y=34 --equation (2)


sub. equation (1) into (2)


7x +5(2x) =34 - equation (3)

7x + 10x = 34

17x = 34
x= 2

sub x= 2 into equation (1)

y = 2x
y= 2(2)
y= 4

check by using the x and y answers in equation (2)

7x+5y=34
7(2) + 5(4) =34

14 + 20 = 34
34 =34

hence answers are ok

maybe I have a different maths system to you PKA ???

 

[pka]

maybe I have a different maths system to you PKA ???

apm, I sorry you misunderstood, but I was not addressing you!
You should read more carefully. You are not at the center of this thread, the poster KP is.