Solve using elimination

Question

y=2/3x+1

2x+3y=27

Answer 1

Our equations from the problem statement are....

y=(2/3)x+1.........Eq1
2x+3y=27...........Eq2

 

Let rewrite Eq1 in standard form Ax+By=C...

-(2/3)x+y=1........Eq1, subtract (2/3)x, both sides

Recommendation: Whenever you have an equation(s) with
fractions in the terms, it is a good idea to write an equivalent
equations with integers only. This greatly reduces the possibility
for making an error in calculations with fractions.

 

-2x+3y=3..........Eq1a, multiply each term by 3

Now our strategy is to eliminate the "y" term by again multiplying
each equation by a factor to get "y" to cancel...

2x- 3y=3...........Eq1a, multiply each term by (-2)

2x+3y=27.........Eq2, as is, no change
------------------
      4x=24.........add each term vertically, "y" term cancels
     ∴ x=6..........divide both sides by 4

 

Substitute 6 for "x" into any of the two ORIGINAL equations
Eq1 or Eq2 and solve for "y." Let's use Eq2...

(2)(6)+3y=27........substitute 6 for "x"
      12+3y=27.......simplify
            3y=15.......subtract 12, both sides
          ∴ y=5.........divide both sides by 3

So our values are x=6 and y=5. Now substitute these values
into the ORIGINAL equations Eq1, Eq2 to check for truth...

5=(2/3)(6)+1........Eq1, substitute values for "x and y"

5=(12/3)+1..........simplify
5=4+1
5=5.....................true, √check

2(6)+3(5)=27.......Eq2, substitute values for "x and y"
     12+15=27.......simplify
           27=27.......true, √check

So the values of x=6 and y=5 were correct.  Always check your
solutions and work!