Help to solve Simultaneous Linear equation by Substitution

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Brinkley23

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Hi All,

New to forum so I do apologise if this is in the wrong section.

I have been given a simultaneous linear equation to solve using the substitution method:

5x - 3y = 26
4x + 2y = 34

Im a bit puzzled and could really do with some help on this as its completely new to me. All I know is that I need to re-arrange? and get a common coefficient by multiplying 1 or both of the equations, but then I'm lost!

Any help would be much appreciated!

Thanks in advance!
 
Brinkley23 said:
Hi All,

New to forum so I do apologise if this is in the wrong section.

I have been given a simultaneous linear equation to solve using the substitution method:

5x - 3y = 26
4x + 2y = 34

Im a bit puzzled and could really do with some help on this as its completely new to me. All I know is that I need to re-arrange? and get a common coefficient by multiplying 1 or both of the equations, but then I'm lost!

Any help would be much appreciated!

Thanks in advance!
Click to expand...
There are several ways to solve set of simultaneous equations. Please show us - mathematically - what do you mean by:

"... I need to re-arrange? and get a common coefficient by multiplying 1 or both of the equations..." applied to this problem.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.
 
Thanks for replying.

Based on the information I have been given to solve this equation (using substitution) I believe that I must do the following:

Equation 1.

5x - 3y = 26

Multiply by 2 resulting in 10x - 6y = 52

Equation 2.

4x + 2y = 34

Multiply by 3 resulting in 12x + 6y = 102

This now gives me the common coefficient for y (6y in both equations)

Im not sure how/if or why i need to rearrange and at what point?

Many Thanks
 
Brinkley23 said:
Thanks for replying.

Based on the information I have been given to solve this equation (using substitution) I believe that I must do the following:

Equation 1.

5x - 3y = 26

Multiply by 2 resulting in 10x - 6y = 52.................................................................(1)

Equation 2.

4x + 2y = 34

Multiply by 3 resulting in 12x + 6y = 102 ...........................................................(2)

This now gives me the common coefficient for y (6y in both equations)

Im not sure how/if or why i need to rearrange and at what point?

Many Thanks
Click to expand...
Equation 1.

5x - 3y = 26

Multiply by 2 resulting in 10x - 6y = 52.................................................................(1)

Equation 2.

4x + 2y = 34

Multiply by 3 resulting in 12x + 6y = 102 ...........................................................(2)

Great work.

Now add (1) and (2) - as defined above.

(10x - 6y) + (12x + 6y) = 52 + 102

10x - 6y + 12x + 6y = 154

22x = 154

x = 154/22 = ?

continue.....
 
You titled this "Help to solve simultaneous linear equations by substitution". What you have done is a perfectly good method but is not "substitution"!

Given the two equations,
5x - 3y = 26
4x + 2y = 34

Solve one of the equations for one unknown in terms of the other. You actually have four choices, solve either of the equations for either unknown. Noticing that all three numbers in the second equation are even, I would divide by 2 to get 2x+ y= 17, then it easy to solve for y= 17- 2x.

Now "substitute" that expression for y in the first equation: 5x- 3y= 5x- 3(17- 2x)= 26.

5x- 51+ 6x= 26
11x- 51= 26
11x= 26+ 51= 77
x= 77/11 which is easier that "154/22" but gives the same answer!

Once you have x, use y= 17- 2x to solve for y.
 
I really appreciate your responses, but I'm just not following (which is really annoying me)

With this equation I was given some material which explains how to use the substitution method, but all methods that people have tried to explain (or that I have looked up) seem to be different. This material in particular that I was given starts off by re-arranging an equation, but doesn't explain why or if you need to apply this to every equation.

Im not after the answer by any means, I really want to be able to understand the steps required to come up with the solution on my own... if at all possible.
 
Brinkley23 said:
I really appreciate your responses, but I'm just not following (which is really annoying me)

With this equation I was given some material which explains how to use the substitution method, but all methods that people have tried to explain (or that I have looked up) seem to be different. This material in particular that I was given starts off by re-arranging an equation, but doesn't explain why or if you need to apply this to every equation.

Im not after the answer by any means, I really want to be able to understand the steps required to come up with the solution on my own... if at all possible.
Click to expand...
This website might be helpful for you:

https://www.purplemath.com/modules/systlin4.htm
 
Brinkley23 said:
I really appreciate your responses, but I'm just not following (which is really annoying me)
Click to expand...
Okay, what can you follow?
Previously I said, "Solve one of the equations for one unknown in terms of the other."
Do you know what that means?
You want to solve equations "by substitution". Do you know what the word "substitute" means?

Here is a simple example,
y= 3x- 4
2x+ 3y= 5

Do you see that you can replace the "y" ("substitute" for y) in the second equation by "3x- 4" because it is equal to y? Do you see that \(\displaystyle 2x+ 3(3x- 4)= 2x+ 3(3x)- 3(4)= 2x+ 9x- 12= 11x- 12\)?

Do you know that 3(3)= 9 and that 3(4)= 12. Do you know that 2x+ 9x= 11x?

Those are pretty elementary but just saying "I don't understand" without showing any work of your own, you haven't given us any idea of what you can do.

With this equation I was given some material which explains how to use the substitution method, but all methods that people have tried to explain (or that I have looked up) seem to be different. This material in particular that I was given starts off by re-arranging an equation, but doesn't explain why or if you need to apply this to every equation.

Im not after the answer by any means, I really want to be able to understand the steps required to come up with the solution on my own... if at all possible.
Click to expand...
 
Subhotosh Khan - Thanks for that link it really helped me.

HallsofIvy - after taking on board your method for finding x, it made it much clearer for me and i think I have come up with a solution;

Full workings:

5x - 3y = 26 .....(1)
4x + 2y = 34 ....(2)

divide equation 2 by 2 resulting in

2x + y = 17

Giving

y = 17 - 2x

Substitute for Y in equation 1

5x - 3y = 5x - 3(17-2x) = 26
5x - 51 + 6x = 26
11x - 51 = 26
11x = 26 + 51 = 77
x = 77/11 = 7

y = 17 - 2x
y = 17 - 2(7)
y = 17 - 14
y = 3

x=7 y=3
 
Brinkley23 said:
Subhotosh Khan - Thanks for that link it really helped me.

HallsofIvy - after taking on board your method for finding x, it made it much clearer for me and i think I have come up with a solution;

Full workings:

5x - 3y = 26 .....(1)
4x + 2y = 34 ....(2)

divide equation 2 by 2 resulting in

2x + y = 17

Giving

y = 17 - 2x

Substitute for Y in equation 1

5x - 3y = 5x - 3(17-2x) = 26
5x - 51 + 6x = 26
11x - 51 = 26
11x = 26 + 51 = 77
x = 77/11 = 7

y = 17 - 2x
y = 17 - 2(7)
y = 17 - 14
y = 3

x=7 y=3
Click to expand...
You got it right !
 
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