x + 3y = -4, 4x + 5y = 5

Question

solve using subsitution

Answer 1

This is a system of equations with two variables.   We want to end up with a solution set that looks like this (x, y) where x an y are numerical values.  This is a simple substitution problem where the goal will be to define our variables x & y such that we can then plug them into the original equation and solve for one of them, then repeat the process and solve for the other.

                                              x + 3y = -4  (Call this Equation 1)

                                             4x + 5y = 5   (Call this Equation 2)

In the equations above, we have two variables in two equations.  The variables are x and y.  We can define the value of the variable x in Equation 1 by subtracting 3y from both sides

                                            x + 3y = -4    (Equation 1)

                                               - 3y     -3y

                                            x = -3y - 4     ("new" Equation 1)

By defining the variable x in Equation 1, I can say, "hey look, now I can put the value of x into the second equation anywhere I find an x.

 

                                               4x + 5y = 5 ( Equation 2)

                                               4(-3y -4) + 5y = 5  (Plug in Equation 1 wherever there is an "x")

                                               -12y -16 + 5y = 5  (distribute the 4 across the binomial)

                                                -7y -16 = 5          (combine like terms)

                                                     +16    +16       (add 16 to both sides)

                                                -7y = 21               (Divide both sides by -7)

                                                    -7

 

                                                 y = -3

 

Now, take this value of y = -3 and plug it back into equation 1 to solve for x.

                                               x + 3y = -4

                                               x + 3(-3) = -4

                                               x -9 = -4                 (Add 9 to both sides)

                                                 +9    +9

                                                   x = 5

The solution set for this system of equations with two variables is (5, -3)

Answer 2

Averie,

  You have two equations in two unknowns.  Use one equation to isolate a parameter, say x, and then substitute that x result into the second equantion to find y.  Finally put the resulting y back into the equation isolating x to evaluate y.  Like so:

Given equations:

x + 3y = -4      (1)

4x + 5y = 5      (2)

From (1) get:

x= -4 -3y          (3)

Substitute this x into (2):

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

-16 -12y +5y = 5

-7y -16 = 5

-7y = 5 +16

y= 21/(-7)

y= -3

Substitute this y value back into  (3):

x= -4 -3y   becomes  x= -4 -3(-3)

x= -4 + 9

x=5

y=-3

 This is an example of a class of problems with multiple equations and multiple unknowns.  The rule is that you need N equations in N unknowns to solve for all N unknowns.  If you have fewer than N equations (say N-1 equations) you can not solve for all the unknowns.  If you have more than N equations, (say N+1), you may end up with contradictory answers for some of the unknowns.

 

BruceS