Substitution Method to solve system of Equations?

Question

Can anyone explain to me how you do the substitution method to solve a system of equations. For example:

-3x+y=-4
2x-y=1

Answer 1

Solve for "x" or "y" in either equation. Let's solve for "y" in the 1st equation..

First: add "3x" to both sides....

- 3x + 3x + y = - 4 + 3x
y = - 4 + 3x

Sec: replace "-4 + 3x" with the y-variable in the 2nd equation....

2x - (- 4 + 3x) = 1
2x - (- 4) - (3x) = 1
2x + 4 - 3x = 1

*Combine "like" terms...

- x + 4 = 1

*Isolate "-x" on one side > subtract "4" from both sides...

- x + 4 - 4 = 1 - 4
- x = - 3

The "x" variable has to be positive - change both signs..

x = 3

Third: replace "3" with the x-variable in the 1st equation...

- 3(3) + y = - 4
- 9 + y = - 4

*Isolate "y" on one side > add "9" to both sides...

- 9 + 9 + y = - 4 + 9
y = 5

Solution: (3, 5)

Answer 2

Solve one of the equations for either x or y and substitute that into the second equation thereby eliminating one of the variables. For example: 2nd equation could become x = y+2. Substituting this into the 1st eqn you get (y +2) +y = -4 Solve this for y and you get y=-3 Therefore x = y+2 = -3+2 = -1

Answer 3

Solve one equation for a variable in terms of another. Lets say in your example you get your 2x-y=1 to 2x-y+y=1+y (note -y+y cancles) then you you divide by 2 on both sides. That produces x=(1+y)/2. Then you put your right side of the equation {(1+y)/2} into your other equation for your x value in the other equation. Thus you get -3[(1+y)/2]+y=4. Now solve for y and then plug your y into the other equation, or you could set both equations equal to one variable such as y. then because your variables are the same then the two modified questions = eachother. Then solve for one variable.
2x-y=1 --> y=-1+2x now set other to y.
-3x+y=-4 --> y=-4+3x now sence y=y
-1+2x=-4+3x now solve for x

Answer 4

in both the equation there r 2 variables 'x' n 'y' unknown...
take any equation say first one

-3x+y = -4 (solve it for any one variable)
-3x = -4 -y (take y on right hand side... i m solving it for x first u can solve it for y also)

3x = 4 + y (taking minus sign commen on both sides)

x = (4 + y) / 3....... (now put this value of x in second equation)

2((4 + y)/3) - y = 1 (solve this one)

(8+2y)/3 - y = 1 (multiplying 2 inside.. now take LCM)

(8 + 2y - 3y) / 3 = 1 (solve it further)

8 -y = 3
y = 8 - 3
y = 5 (now put this value in any of the above given equations to take out value of x)

taking second equation

2x - y = 1
2x - 5 = 1
2x = 5 + 1
2x = 6
x = 6/2
x = 3

now u got the value of x n y both
do a verification put the value of x n y in the remaning equation dat is first one

-3x + y = -4
-3(3) + 5 = -4
-9 + 5 = -4
-4 = -4

left hand side = Right hand side means these valus of x n y satisfies both equation means they r correct.

Answer 5

solve the first equation for y bu adding 3x to both sides:
y = 3x - 4

substitute this into the second equation for y and solve for x:
2x - (3x - 4) = 1
-x + 4 = 1
-x = -3
x = 3

substitute back into the first equation after solving for y:
y = 3x - 4
y = 3(3) - 4
y = 5

Answer 6

Take one equation and get it so that one variable is alone. I would make the top one y=3x-4, then in the bottom equation put 3x-4 in for y. 2x-(3x-4)=1. Then solve, and after you get x, plug that answer in for x in either equation and solve for y.

Answer 7

-3x+y=-4 add 3x to each side
y=3x-4


2x-y=1 substitute 3x-4 for y
2x-(3x-4)=1
2x-3x+4=1
-x+4=1 subtract 4 from each side
-x=-3 multiply both sides by -1
x=3
y=3x-4 substitute 3 for x
y=3*3-4
y=9-4
y=5

(3, 5)

Without being told to use substitution, I would use addition
-3x+y=-4
2x-y=1 add
-x=-3
x=3 & substitute 3 for x in either equation to find y.

Only a poor teacher would tell students to use substitution for this problem.