Basic Algebra/Solving Equations/Equations with More than One Variable

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Sometimes, there will be more than one variable that needs to be solved for in an equation. There are systems used by mathematicians to solve for these unknown quantities, which are discussed below.

Vocabulary[edit]

Variable - a letter (A-Z) that takes the place of a number.

Expression - a group of numbers and variables, added, multiplied, subtracted or divided by each other

(examples include: 2,  7x,  8x-7,  \sqrt{x} + 4 )

Equation - an expression that is equal to another expression

Coefficient - the number multiplied by a variable (the 3 in 3x)

Lesson[edit]

There are three simple ways to solve an equation with two variables.

The number of variables in an equation corresponds to the number of equations needed to solve for the variables.

Graphing[edit]

The most simple way to solve for the variables is by graphing both equations and finding the point where they intersect. This method is not incredibly accurate, unless we measure it with a ruler however, since we cannot be sure exactly where an intersection is unless we measure it.

The substitution method[edit]

In these, the first step is isolating one variable on one side of an equation. After this is accomplished, the expression on the other side of the equal sign can be substituted for the variable we solved for in the first step. Now, this second equation has only one variable. We simplify the equation, solving for the remaining variable. Then we plug this value back into one of the original equations and solve for the first variable.

Elimination method[edit]

To use the elimination method, we make the coefficients of one variable opposites of each other by multiplying or dividing both sides of the equations by the desired numbers. Once this is accomplished, we arrange the equations so that each variable is above the same variable in the other equation. After this, we add the coefficients of each variable to each other to produce a new equation, with only one variable (since the coefficients of one set of variables will cancel. Then follow the steps listed under substitution to solve for the remaining variable(s). (Refer to example 3 to see this method in action)

Example Problems[edit]

Example 1[edit]

3x + 2 = y
4x - 6 = 2y
  • Substitute (3x + 2) for y in the second equation
4x - 6 = 2(3x + 2)
  • Distribute the 2
 4x - 6 = 6x + 4
  • Add 6 to each side
 4x = 6x + 10
  • Subtract 6x from each side
 -2x = 10
  • Divide each side by -2
x = -5
  • Now, plug -5 in for x in one of the original equations and solve for y:
3(-5) + 2 = y
  • Simplify
 -15 + 2 = y
 -13 = y

So the solutions are x=-5 and y=-13 (If you graphed these lines, the intersection would be at the point (-5,-13)

Example 2[edit]

  •  5x+3y=12
  •  x - y=4
  • Multiply the second equation by 3 on each side so that the coefficients of the y's will be 3 and -3
  •  5x+3y=12
  •  3x - 3y=12
  • "Add" the equations together using the addition method
  •  8x + 0y = 24
  • Divide each side of the equation by 8
  •  x=3
  • Plug this 3 back into the first equation for x and solve for y
  •  (3) - y = 4
  •  -y = 1
  •  y = -1

So, x=3 and y= -1

Example 3[edit]

2y = 3x + 20
4y = 6x + 24
  • Divide the equations by 2 and 4, respectively
y = \frac{3}{2}x + 10
y = \frac{3}{2}x + 6
  • Substitute
\frac{3}{2}x + 10 = \frac{3}{2}x + 6
  • Subtract \frac{3}{2}x from each side
 10 = 6

This absurd equation is obviously false, therefore there are no solutions. This is because the coefficients of the x were the same while in slope-intercept form, so the lines were parallel (never touching) so there are no solutions. (For a further explanation of slope-intercept form, see Slope-Intercept Form of a Line)

NOTE: When you get and equation that is always true (3=3 or 5=5) there are infinitely many solutions.

Practice Games[edit]

There is a 2-variable game on the following webpage: [[1]]

Practice Problems[edit]

Problem 1[edit]

  • y=x + 2
  • y=2x - 1

x=3, y=5

Problem 2[edit]

  • x + y = 5
  • 2x - y = 10

x=5, y=0

Problem 3[edit]

  • 4x + 6 = 2y
  • x + y=5

x=\frac{2}{3}, y=\frac{13}{3}

Problem 4[edit]

  • x - 2y=7
  • y - x=13

x= -33, y= -20

Problem 5[edit]

  • y=2x - 1
  • 2y + 2= 4x

Infinitely many solutions