3x-5y=1
9x+20y=10
system of elimination
please explain
9x+20y=10
system of elimination
please explain
-
Equation A: 3x - 5y = 1
Equation B: 9x + 20y = 10
Solving the systems of equations with "substitution" would be difficult so I will solve the equations with elimination. When I perform elimination I am allowed to...
(1) Multiply either [or even both equations] by some number so as to eliminate either variable "x" or variable "y". For example, the solution to...
2x + 5 = 9 is "x" = 2. If I multiply the whole equation by 2, for example, I get this:
4x + 10 = 18, and the solution is still "x" = 2. I simply scaled up the measurements. It works the same way for systems of equations.
(2) Equations can be either "subtracted" or "added" to eliminate a variable.
-----------------------------------
To eliminate "x" multiply equation A by 3
(Equation A)(3) -------> 3[ 3x - 5y = 1] ---------> 9x - 15y = 3
Equation B [ as is] -----------------------------------> 9x + 20y = 10
"Subtract" the systems of equations. Subtract!!! Not add!! not this time!
"9x - 9x" gives me no more "x's", "-15y - 20y" gives me "-35y" and "3 - 10" gives me "-7"
So, I have this:
-35Y = -7
-35(Y) / -35 = -7 / -35
Y = 1 / 5 or 0.2
Now, substitute that "Y" value into either equation. I'm in the mood for equation A:
3x - 5(1/5) = 1
3x - 1 = 1
3x - 1 [+1] = 1 [+1]
3x = 2
3(x) / 3 = 2 / 3
x = 2/3 ~ 0.67
Solution set: x = 2 / 3 ; Y = 1 / 5
I don't usually make mistakes with these kinds of problems. But, I still suggest that you check your work each time...in case I deceived you [which I did not].
Equation B: 9x + 20y = 10
Solving the systems of equations with "substitution" would be difficult so I will solve the equations with elimination. When I perform elimination I am allowed to...
(1) Multiply either [or even both equations] by some number so as to eliminate either variable "x" or variable "y". For example, the solution to...
2x + 5 = 9 is "x" = 2. If I multiply the whole equation by 2, for example, I get this:
4x + 10 = 18, and the solution is still "x" = 2. I simply scaled up the measurements. It works the same way for systems of equations.
(2) Equations can be either "subtracted" or "added" to eliminate a variable.
-----------------------------------
To eliminate "x" multiply equation A by 3
(Equation A)(3) -------> 3[ 3x - 5y = 1] ---------> 9x - 15y = 3
Equation B [ as is] -----------------------------------> 9x + 20y = 10
"Subtract" the systems of equations. Subtract!!! Not add!! not this time!
"9x - 9x" gives me no more "x's", "-15y - 20y" gives me "-35y" and "3 - 10" gives me "-7"
So, I have this:
-35Y = -7
-35(Y) / -35 = -7 / -35
Y = 1 / 5 or 0.2
Now, substitute that "Y" value into either equation. I'm in the mood for equation A:
3x - 5(1/5) = 1
3x - 1 = 1
3x - 1 [+1] = 1 [+1]
3x = 2
3(x) / 3 = 2 / 3
x = 2/3 ~ 0.67
Solution set: x = 2 / 3 ; Y = 1 / 5
I don't usually make mistakes with these kinds of problems. But, I still suggest that you check your work each time...in case I deceived you [which I did not].