Simultaneous equations in word form

[CSmith1]

Two numbers are such that 5 times the first is three more than 6 times more than the second. Four times the second is five more than the first. Find the numbers.

and also The sum of two nubers is 90. If 20 is added to 3 times the smaller number, the result exceeds twice the larger by 50. Find the numbers.

 

[MarkFL]

Re: HELp with simultaneous equations in word form

First, let me say this is not a differential equations topic.

1.) Let x be the first number and y be the second number. The statement:

"5 times the first is three more than 6 times the second"

allows us to write:

5x = 6y + 3

The statement:

"Four times the second is five more than the first"

allows us to write:

4y = x + 5

I am assuming your difficulty is in setting up the equations and not is actually solving a 2X2 linear system. Can you proceed from here?

2.) Let x be the first number and y be the second number. The statement:

"The sum of two numbers is 90"

allows us to write:

x + y = 90

The statement:

"If 20 is added to 3 times the smaller number, the result exceeds twice the larger by 50"

If we decide to let x < y then we may write:

3x + 20 = 2y - 50

which we may arrange as:

3x - 2y = -70

Can you proceed from here?

 

[CSmith1]

Re: HELp with simultaneous equations in word form

okay so for the first question would it be lined up like

5x-6y=3
x-4y=5

and then i solve as a simultaneous equation?

 

[MarkFL]

Re: HELp with simultaneous equations in word form

You want -5 on the right side of your second equation. Then yes, solve the simultaneous system.

 

[CSmith1]

4y = x + 5

but if i bring 4y over to the equal sign it becomes x-4y and 5 is already behind the sign. so how would it become -5

 

[Jameson]

Hi CSmith,

As MarkFL said you start with $4y = x + 5$. You're looking to get both variables on one side of the equation and the constant on the other side to match the form of your first equation. There are a couple of ways to do this, both equally valid.

$4y = x + 5$
$4y-x=5$ (1)

$4y = x + 5$
$0=x+5-4y$
$-5=x-4y$
$x-4y=-5$ (2)

(1) and (2) are really the same. If you take either one, multiply every term by -1 then you'll get the other.

 

[MarkFL]

4y = x + 5

If we subtract 4y + 5 from both sides:

4y - (4y + 5) = x + 5 - (4y + 5)

-5 = x - 4y

 

[CSmith1]

so how would i line up the first question to solve it?

 

[MarkFL]

One way would be to use elimination. We have:

5x - 6y = 3

x - 4y = -5

To eliminate y, we could multiply the first equation by 2 and the second by -3:

10x - 12y = 6

-3x + 12y = 15

Now add the equations, and you will have a linear equation in x only. Then, once you have x, you may substitute this value into either of you original equations to determine y.

 

[CSmith1]

how would u know to multiply the first equation by 2 and the second by -3 ?

thank u so much for ur help

 

[MarkFL]

By observing that the LCM of 4 and 6 is 12. You could also eliminate x by multiplying the second equation by -5, since the LCM of 1 and 5 is 5. Do you see how this works?