SOLUTION: Solve system by substitution or addition, whichever is easier.
2y - x = 3
x = 3y - 5
Algebra.Com
Question 134244: Solve system by substitution or addition, whichever is easier.
2y - x = 3
x = 3y - 5
Answer by algebrapro18(249) (Show Source): You can put this solution on YOUR website!
Since one of the equations has a variable solved for already, lets use substitution.
The steps in substitution are as follows:
1) Solve one of the equations for a variable.
2) Plug in the value of that variable into the other equation and solve for the one variable left.
3) Plug that value into 1 and solve for the first variable.
Since step one has already been done for us we can just jump to step to.
2y - x = 3
x = 3y - 5
Plugging x = 3y-5 into the top equation you get:
2y -(3y-5) = 3
now we just solve for y
2y -(3y-5) = 3
2y - 3y + 5 = 3
-y + 5 = 3
-y = -2
y = 2
Now we move on to step 3. Plugging the value y = 2 into the second equation we get:
x = 3(2)-5
x = 6 - 5
x = 1
now we are done, our solution is (1,2) or x = 1 and y =2 depending on what form we want to leave it in.
RELATED QUESTIONS
Solve each system by substitution or addition, whichever is easier.
52. 2y – x = 3
(answered by jim_thompson5910)
Solve each system by substitution or addition, whichever is easier.
52. 2y - x = 3
(answered by ewatrrr)
I need to solve this by substitution or addition, whichever one is easier.
2y-x=3... (answered by KnightOwlTutor)
I need to solve by substitution or addition, whichever one is easier.
2y - x = 3
x =... (answered by solver91311)
I need to solve by substitution or addition, whichever one is easier.
2y - x = 3
x = (answered by solver91311)
50. Solve each system by substitution or addition, whichever is easier.
2y-x=3... (answered by doukungfoo)
Solve each system by substitution or addition, whichever is easier.
2y-3x=-1
5y+3x=29
(answered by richard1234)
I need help to solve by substitution or addition, whichever one is easier. Please show... (answered by mananth)
Hi, I need help with the following:
1. Solve by using substitution method.... (answered by checkley71)