Chapter 1 Linear Equations
1.1 Use a General Strategy to Solve Linear Equations
Learning Objectives
By the end of this section, you will be able to:
- Solve linear equations using a general strategy
- Classify equations
- Solve equations with fraction or decimal coefficients
Solve Linear Equations Using a General Strategy
Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!
Solution of an Equation
A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.
To determine whether a number is a solution to an equation, we substitute the value for the variable in the equation. If the resulting equation is a true statement, then the number is a solution of the equation. If the resulting equation is not true, then the number is not a solution of the equation.
Example 1
Determine whether the values are solutions to the equation [latex]5y+3=10y-4[/latex].
a) [latex]y=\frac{3}{5}[/latex]
Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.
[latex]\begin{array}{lrcl} & 5y+3 & = & 10y-4\\ \text{Substitute} {\frac{3}{5}} \text{for y.} & 5({\frac{3}{5}})+3 & \stackrel{?}{=} & 10({\frac{3}{5}})-4 \\ \text{Multiply.} & 3+3 & \stackrel{?}{=} & 6-4\\ \text{Simplify.} & 6 & \neq & 2\\ \end{array}[/latex]
Since [latex]y=\frac{3}{5}[/latex] does not result in a true equation, [latex]y=\frac{3}{5}[/latex] is not a solution to the equation [latex]5y+3=10y-4[/latex]
b) [latex]y=\frac{7}{5}[/latex]
Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.
[latex]\begin{array}{lrcl} & 5y+3 & = & 10y-4\\ \text{Substitute} {\frac{7}{5}} \text{for y.} & 5({\frac{7}{5}})+3 & \stackrel{?}{=} & 10 ({\frac{7}{5}})-4 \\ \text{Multiply.} & 7+3 & \stackrel{?}{=} & 14-4\\ \text{Simplify.} & 10 & = & 10\\ \end{array}[/latex]
Since [latex]y=\frac{7}{5}[/latex] results in a true equation, [latex]y=\frac{7}{5}[/latex] is a solution to the equation [latex]5y+3=10y-4[/latex]
Exercise 1
Determine whether the values are solutions to the equation: [latex]9y+2=6y+3[/latex]
a) [latex]y=\frac{4}{3}[/latex]
b) [latex]y=\frac{1}{3}[/latex]
Solution
a) no
b) yes
Determine whether the values are solutions to the equation: [latex]4x-2=2x+1[/latex]
a) [latex]x=\frac{3}{2}[/latex]
b) [latex]x=-\frac{1}{2}[/latex]
Solution
a) yes
b) no
There are many types of equations that we will learn to solve. In this section we will focus on a linear equation.
A linear equation is an equation in one variable that can be written as [latex]ax+b=0[/latex], where [latex]a[/latex] and [latex]b[/latex] are real numbers and [l