linear equations, math methods, elimination method

[TMAS]

what determines the number used to divide or multiply in the elimination method??? for eg 6S + 4C = 58
5S + 2C = 35

 

[Deleted member 4993]

what determines the number used to divide or multiply in the elimination method??? for eg 6S + 4C = 58
5S + 2C = 35

That depends on which variable (S or C), you want to eliminate.

 

[lev888]

what determines the number used to divide or multiply in the elimination method??? for eg 6S + 4C = 58
5S + 2C = 35

Multiply the second equation by 2 to make the C terms the same (4C). Now you can subtract the equations to eliminate C.
Can you figure out how to eliminate S instead?

 

[Dr.Peterson]

what determines the number used to divide or multiply in the elimination method??? for eg 6S + 4C = 58
5S + 2C = 35

In general, you choose which variable to eliminate, find the LCM of the two coefficients of that variable, and choose multipliers to obtain that LCM from each of the numbers (with opposite signs if the goal is to add the resulting equations.

In the case of C, as lev888 mentioned, the LCM of 4 and 2 is 4, and we multiply by 1 and by 2 respectively (or -1 and 2 so we can add). It's a little harder to eliminate C, so I would do that only if it were hard to find S from the value we find for C.

 

[eddy2017]

Multiply the second equation by 2 to make the C terms the same (4C). Now you can subtract the equations to eliminate C.
Can you figure out how to eliminate S instead?

if the second term is alike you're home free.
[math]6S + 4C = 58[/math][math]10s + 4c =70[/math]

 

[eddy2017]

you can eliminate c easily.
find the value of c by subtracting the two equations.
it will do you a world of good to multiply the first equation by 6
it will do you a whole universe of good to multiply the second equation by 5 ( you need to try your best to get two terms to cancel out when you do the sub)
[math]5(6s + 4c)=5(58)[/math][math]6(5s + 4c)=6(35)[/math]distributing
[math]30s + 20c= 290[/math][math]30s + 24c =210[/math]
now you subtract these two guys vertically ( imagine there is a minus sign to the utmost left
[math]0 - 4c =80[/math][math]-4c=80[/math][math]c= 80/-4[/math][math]c=-20[/math]you have the value of c now

can you find the value of s now?

the tutors will give way less than what I have given you here, so count your lucky stars!

 

[topsquark]

if the second term is alike you're home free.
[math]6S + 4C = 58[/math][math]10s + 4c =70[/math]

Just a note: In general the variable C and the variable c represent two different things. Mathematics is "case sensitive!"

-Dan

 

[eddy2017]

Just a note: In general the variable C and the variable c represent two different things. Mathematics is "case sensitive!"

-Dan

Good to know!. Thank you. It was a typo, though.

 

[eddy2017]

Good to know!. Thank you. It was a typo, though

The set of complex numbers is represented by the Latin capital letter C.
The small c for variables and coefficient but spot-on remark. That keeps me on my toes. Lol

 

[Deleted member 4993]

The set of complex numbers is represented by the Latin capital letter C.
The small c for variables and coefficient but spot-on remark. That keeps me on my toes. Lol

The point of Response #7 is that:

you CANNOT use two different symbols ( C & c) to describe the SAME (one) VARIABLE in the SAME problem.​

 

[Deleted member 4993]

you can eliminate c easily.
find the value of C by subtracting the two equations.
it will do you a world of good to multiply the first equation by 6

That process would be numerically correct but unnecessarily lengthy and time consuming.

Replace the value of 'C' in one of the equations [and the 2 given equations with 2 unknowns will be reduced to one equation with one unknown - namely 'S'] and solve for 'S'.

 

[eddy2017]

That process would be numerically correct but unnecessarily lengthy and time consuming.

Replace the value of 'C' in one of the equations [and the 2 given equations with 2 unknowns will be reduced to one equation with one unknown - namely 'S'] and solve for 'S'.

I said the same exact thing ( maybe in other words) back at #6

 

[Deleted member 4993]

I said the same exact thing ( maybe in other words) back at #6

You better check your work in response #6

The correct value of 'C' should be 10

Consequently,

6*S + 4*10 = 58 → 6*S = 18 → S = 3

S = 3​

C = 10​

​

Now check answer (very important step)

6S + 4C = 58
5S + 2C = 35

6 * 3 + 4*10 = 58.........................√

5*3 + 2*10 = 35 ..........................√

 

[eddy2017]

lev said : 'Multiply the second equation by 2 to make the C terms the same (4C).
that is what I did and that is how I got. I multiply THE original second equations by 2 and that is how I got 10s + 4c=70
6S+4C=58
10s+4c=70

 

[Dr.Peterson]

The problem was

what determines the number used to divide or multiply in the elimination method??? for eg 6S + 4C = 58
5S + 2C = 35

You appear to be looking at #5. What you said in #6 was

[math]5(6s + 4c)=5(58)[/math][math]6({\color{Red}\mathbf{5s + 4c}})=6(35)[/math]...
[math]{\color{Red}\mathbf{c=-20}}[/math]you have the value of c now

So you just copied a number incorrectly, which resulted in a wrong answer.

That's why checking (in the original equation) is important!

 

[eddy2017]

The problem was

You appear to be looking at #5. What you said in #6 was

So you just copied a number incorrectly, which resulted in a wrong answer.

That's why checking (in the original equation) is important!

Yes, I see it now. Thanks!.