Lesson 2.1: Use a General Strategy to Solve Linear Equations

New Concept

This lesson introduces a powerful, step-by-step strategy for solving any linear equation. You'll learn to simplify expressions, isolate variables, and confidently handle equations with fractions or decimals to find the solution.

What’s next

Next, you’ll see this strategy in action with interactive examples, followed by practice cards to build your equation-solving skills.

Property

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:

Step 1. Substitute the number for the variable in the equation.
Step 2. Simplify the expressions on both sides of the equation.
Step 3. Determine whether the resulting equation is true.

• If it is true, the number is a solution.
• If it is not true, the number is not a solution.

Examples

• Is $x=5$ a solution to the equation $3x - 5 = 10$? We substitute $3(5) - 5 = 15 - 5 = 10$. Since $10 = 10$, it is a solution.
• Is $y=-2$ a solution to the equation $4y + 9 = 2y$? We substitute $4(-2) + 9 = -8 + 9 = 1$ on the left, and $2(-2) = -4$ on the right. Since $1 \neq -4$, it is not a solution.
• Is $a = \frac{1}{3}$ a solution to the equation $9a + 2 = 5$? We substitute $9(\frac{1}{3}) + 2 = 3 + 2 = 5$. Since $5=5$, it is a solution.

Property

A linear equation is an equation in one variable that can be written as $ax + b = 0$, where $a$ and $b$ are real numbers and $a \neq 0$.
To solve linear equations using a general strategy:
Step 1. Simplify each side of the equation as much as possible by using the Distributive Property and combining like terms.
Step 2. Collect all variable terms on one side of the equation.
Step 3. Collect all constant terms on the other side.
Step 4. Make the coefficient of the variable term equal to 1.
Step 5. Check the solution by substituting it into the original equation.

Examples

• Solve $5(x-3) + 1 = -9$. First, distribute: $5x - 15 + 1 = -9$. Combine terms: $5x - 14 = -9$. Add 14 to both sides: $5x = 5$. Divide by 5 to get $x=1$.
• Solve $8k - 5 = 2k + 13$. Subtract $2k$ from both sides: $6k - 5 = 13$. Add 5 to both sides: $6k = 18$. Divide by 6 to get $k=3$.
• Solve $-(z+4) = 10$. Distribute the negative sign: $-z - 4 = 10$. Add 4 to both sides: $-z = 14$. Multiply by $-1$ to get $z=-14$.

Explanation

The main goal is to isolate the variable. Think of it as unwrapping a gift: use inverse operations in reverse order to get the variable by itself. Always simplify both sides first to make the process easier.

Property

Equations can be classified into three types based on their solutions:

• A conditional equation is true for one or more specific values of the variable.
• An identity is an equation that is true for all real numbers. Solving results in a true statement, such as $3=3$.
• A contradiction is an equation that has no solution. Solving results in a false statement, such as $5=-2$.

Examples

• Conditional: $4x-1=7$. Solving this gives $4x=8$, so $x=2$. The equation is true only for this value.
• Identity: $2(x+3) = 2x+6$. Distributing gives $2x+6=2x+6$. Subtracting $2x$ from both sides results in $6=6$. This is an identity, and the solution is all real numbers.
• Contradiction: $3y = 3(y-2)$. Distributing gives $3y = 3y-6$. Subtracting $3y$ from both sides results in $0=-6$. This is a contradiction and has no solution.

Explanation

Not all equations have just one answer. If solving leads to a specific value like $x=2$, it's conditional. If the variables cancel and you get a true statement like $5=5$, it's an identity. If you get a false statement, it's a contradiction.

Property

To solve equations with fraction or decimal coefficients:
Step 1. Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
Step 2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
Step 3. Solve using the General Strategy for Solving Linear Equations.

Examples

• Solve $\frac{1}{5}x - 1 = \frac{3}{5}$. The LCD is 5. Multiply all terms by 5: $5(\frac{1}{5}x) - 5(1) = 5(\frac{3}{5})$, which simplifies to $x - 5 = 3$. The solution is $x=8$.
• Solve $0.2y + 0.05 = 0.45$. To clear decimals, multiply by 100: $100(0.2y) + 100(0.05) = 100(0.45)$, which gives $20y + 5 = 45$. This simplifies to $20y = 40$, so $y=2$.
• Solve $\frac{1}{2}(z+1) = \frac{2}{3}$. The LCD is 6. Multiply both sides by 6: $6 \cdot \frac{1}{2}(z+1) = 6 \cdot \frac{2}{3}$, which gives $3(z+1) = 4$. So, $3z+3=4$, which means $3z=1$, and $z=\frac{1}{3}$.

Explanation

Fractions and decimals can be messy. To simplify your work, find the least common denominator (LCD) of all terms and multiply both sides of the equation by it. This clears the fractions, leaving a simpler integer equation to solve.