That depends on which variable (S or C), you want to eliminate.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.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.what determines the number used to divide or multiply in the elimination method??? for eg 6S + 4C = 58
5S + 2C = 35
if the second term is alike you're home free.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?
Just a note: In general the variable C and the variable c represent two different things. Mathematics is "case sensitive!"if the second term is alike you're home free.
[math]6S + 4C = 58[/math][math]10s + 4c =70[/math]
Good to know!. Thank you. It was a typo, though.Just a note: In general the variable C and the variable c represent two different things. Mathematics is "case sensitive!"
-Dan
The set of complex numbers is represented by the Latin capital letter C.Good to know!. Thank you. It was a typo, though
The point of Response #7 is that: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
That process would be numerically correct but unnecessarily lengthy and time consuming.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
I said the same exact thing ( maybe in other words) back at #6That 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'.
You better check your work in response #6I said the same exact thing ( maybe in other words) back at #6
6 * 3 + 4*10 = 58.........................√6S + 4C = 58
5S + 2C = 35
You appear to be looking at #5. What you said in #6 waswhat determines the number used to divide or multiply in the elimination method??? for eg 6S + 4C = 58
5S + 2C = 35
So you just copied a number incorrectly, which resulted in a wrong answer.[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
Yes, I see it now. Thanks!.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!