Unit 5 - Objective 4 - Solving 2 equations in 2 unknowns


Graphical methods are not very accurate ways to determine the common solutions of simultaneous equations.  We will look at 2 algebraic methods for solving 2 equations in 2 unknowns.

Solution by Substitution

 

Example:

Solve the system

2x - 3y = 2
3x + 2y = 3

We could solve the first equation for y.

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Substitute this solution for y in the other equation.

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Solve this equation for x.

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We then substitute this solution for x into the first equation.

2 (1) - 3y = 2
2 - 3y = 2
- 3y = 0
y = 0

The solution is (1, 0) which was the same answer from objective 3.

 

Solution by Addition or Subtraction

Add or subtract the two equations in order to eliminate one of the variables.  It is sometimes necessary to multiply the original equation by a constant before you can eliminate one of the variables by adding or subtracting the equations.

 

Examples:

2x - 3y = 2
3x + 2y = 3

We could multiply equation (1) by 3.

6x - 9y = 6

We could multiply equation (2) by -2.

-6x - 4y = -6

We can now add the equations.

6x - 9y = 6
- 6x - 4y = - 6

0x - 13y = 0
y = 0

We can substitute this value into either equation to find x.

6x - 9 (0) = 6
6x = 6
x = 1

So (1, 0) is the solution which satisfies these two linear equations.

 

The following is an example of a verbal problem by first setting up the proper equations and then solving.

 

Examples:

How much of a solution that is 50% sulfuric acid should be mixed to a solution that is 10% sulfuric acid to obtain 20 gallons of a mixture that is 20% sulfuric acid?

Solution

Write down the meaning of the variables

let x = the number of gallons of 50% solution needed
let y = the number of gallons of 20% solution needed

Write an equation for the total quantity, in gallons

x + y = 20  (x gallons + y gallons = 20 gallons)

Write and equation for the amount of sulfuric acid, the concentration.

(50% of 1st solution) + (10% of 2nd solution) = (20% of total solution)
(0.50) (x gallons) + (0.10) (y gallons) = (0.20) (20 gallons)
0.50x + 0.10y = 0.20 (20)
or
0.5x + 0.1y = 4

So we need to solve

x + y = 20
0.5x + 0.1y = 4

If we want to eliminate the decimals, you can multiply the second equation by 10 and it would become 5x + y = 40.  So we have

x + y = 20
5x + y = 40

We could multiply the second equation by - 1 and add to the first equation.

x + y = 20
- 5x - y = - 40

- 4x + 0y = - 20
x = 5

x = 5 gallons of the 50% solution

If we substitute x = 5 into the first equation,

5 + y = 20
y = 15 gallons of the 10% solution


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