Algebra 1 help please?

So in school we started Solving Systems Using Substitution. And I'm having a little trouble understanding so could anyone show how you solve any of these problems step by step. (If possible all problems please) And if possible explain why you do what you do in each step. Thanks

1. 3x+3y=3

-5x+5y=-85

2. -x+y=-3

-x-y=-15

3. y=-2+0

y+3x=-1

4. x=-5y-9

-3x-2y=14

5. 2x=-5y-24

x=-4y-18

Update:

My book says Substitution method replacing one variable with an equivalent expression containing the other variable, you can make a one-variable equation.

Steps

1. Select one equation and solve for one of its variables

2. In the other equation substitute for the variable you just solved

3. Solve new equation

4. Back substitute to solve for the remaining variables

I looked at the examples and tried to solve problems but I still don't quite understand the whole thing.

3 Answers

Relevance
  • 8 years ago
    Favorite Answer

    1.) 3x + 3y = 3

    3y = - 3x + 3

    y = - x + 1........................... Eq. 1

    - 5x + 5y = - 85 ................. Eq. 2

    Sub - x + 1 from Eq. 1 for y in Eq. 2:

    - 5x + 5(- x + 1) = - 85

    - 5x - 5x + 5 = - 85

    - 10x = - 85 - 5

    - 10x = - 90

    x = - 90 / - 10

    x = 9

    and

    y = - 9 + 1

    y = - 8

    Solution Set (9, - 8)

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    2.) - x + y = - 3

    y = x - 3 ............................. Eq. 1

    - x - y = - 15 ...................... Eq. 2

    Sub x - 3 for y in Eq. 2:

    - x - (x - 3) = - 15

    - x - x + 3 = - 15

    - 2x = - 15 - 3

    - 2x = - 18

    x = - 18 / - 2

    x = 9

    and

    y = 9 - 3

    y = 6

    Solution Set (9, 6)

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    3.) y = - 2 + 0

    y = - 2 ............................... Eq. 1

    y + 3x = - 1 ........................ Eq. 2

    Sub - 2 for y in Eq. 2:

    - 2 + 3x = - 1

    3x = 2 - 1

    3x = 1

    x = 1/3

    Solution Set (1/3, - 2)

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    4.) x = - 5y - 9 ................... Eq. 1

    - 3x - 2y = 14 .................... Eq. 2

    Sub - 5y - 9 for x in Eq. 2:

    - 3(- 5y - 9) - 2y = 14

    15y + 27 - 2y = 14

    13y = 14 - 27

    13y = - 13

    y = - 13 / 13

    y = - 1

    and

    x = - 5(- 1) - 9

    x = 5 - 9

    x = - 4

    Solution Set (- 4, - 1)

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    5.) 2x = - 5y - 24

    x = - 5/2 y - 12 .................. Eq. 1

    x = - 4y - 18 ...................... Eq. 2

    Sub - 5/2 y - 12 for x in Eq. 2:

    - 5/2 y - 12 = - 4y - 18

    2(- 5/2 y) - 2(12) = 2(- 4y) - 2(18)

    - 5y - 24 = - 8y - 36

    - 5y + 8y = 24 - 36

    3y = - 12

    y = - 12 / 3

    y = - 4

    and

    x = - 4(- 4) - 18

    x = 16 - 18

    x = - 2

    Solution Set (- 2, - 4)

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

     

    Source(s): 11/29/12
  • Anonymous
    8 years ago

    Pick either "x" or "y" to solve for, in any of the equation,

    Let's try to solve for x in the first equation:

    1. 3x+3y=3 ---> First Equation

    -5x+5y=-85

    > 3x + 3y = 3

    > 3x = 3 - 3y

    Divide both sides of the equation by "3" to get x:

    >3x/3 = 3/3 - 3y/3

    x = 1 - y ---> This is what you are going to use to substitute in the second equation:

    3x+3y=3

    -5x+5y=-85 ---> Second Equation

    Substitute the solved value of x from the first equation to the variable "x" in the second equation:

    So:

    - 5(1 - y) + 5y = - 85

    Simplify:

    > - 5 + 5y + 5y = -85

    > - 5 + 10y = - 85

    > 10y = - 85 + 5

    > 10y = -80, divide both sides by 10 do get "y"

    > 10y/10 = -80/10

    > y = -8, now you have a definite value of "y", just substitute this to any of the given equations:

    3x+3y=3

    -5x+5y=-85

    Let's choose 3x + 3y = 3;

    So just substitute the value of y to the equation:

    > 3x + 3(-8) = 3

    > 3x - 27 = 3

    > 3x = 3 + 27

    > 3x = 30, divide both sides of the equation by "3" to get x:

    > 3x/3 = 30/3

    > x = 10, now you have a definite value of x:

    So x = 10, y = -8

    2. Same steps as the first one:

    -x+y=-3 ---> First Equation, solve for x:

    -x-y=-15

    > - x + y = - 3

    > - x = - 3 - y

    > - x = - 3 - y, divide both sides by "-1" to get x

    > - x/-1 = - 3/-1 - y/-1

    > x = 3 + y

    -x+y=-3

    -x-y=-15 ---> Second equation, substitute the value of x from the first equation:

    So:

    > - (3 + y) - y = - 15

    > - 3 - y - y = - 15

    > - 3 - 2y = - 15

    > - 2y = - 15 + 3

    > - 2y = -12, divide both sides by "-2" to get y:

    > -2y/-2 = -12/-2

    > y = 6, now you have a definite value of y, substitute to any of the given equation:

    -x+y=-3

    -x-y=-15 ---> Let's choose this:

    > - x - (6) = - 15

    > - x = - 15 + 6

    > - x = - 9, divide both sides by "-1" to get x:

    > - x/ -1 = - 9/-1

    > x = 9

    So x = 9, y = 6

    3. y=-2+0, you don't need to solve this y = -2

    y+3x=-1

    So (- 2) + 3x = -1

    > - 2 + 3x = - 1

    3x = - 1 + 2

    3x = 1

    3x/3 = 1/3

    x = 1/3

    Well you get the idea, just follow the steps.

  • 8 years ago
    Source(s): sophomore in high school :)
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