Lesson 3: Algebraic Solution of Systems

New Concept

This lesson moves beyond graphing to find precise solutions for systems of linear equations. You will master two powerful algebraic methods: substitution and elimination, and learn to apply them to real-world problems.

What’s next

Now, you'll dive into the substitution method with worked examples and practice cards to build your skills step-by-step.

Property

To Solve a System by Substitution.

1. Choose one of the variables in one of the equations. (It is best to choose a variable whose coefficient is 1 or -1.) Solve the equation for that variable.
2. Substitute the result of Step 1 into the other equation. This gives an equation in one variable.
3. Solve the equation obtained in Step 2. This gives the solution value for one of the variables. Substitute this value into the result of Step 1 to find the solution value of the other variable.

Examples

• To solve the system $y = 4x + 2$ and $y = x + 8$, set the expressions for $y$ equal: $4x + 2 = x + 8$. Solving gives $3x = 6$, so $x=2$. Substitute $x=2$ into the second equation to get $y = 2 + 8 = 10$. The solution is $(2, 10)$.
• To solve the system $x - 3y = 7$ and $2x + y = 7$, first solve the second equation for $y$ to get $y = 7 - 2x$. Substitute this into the first equation: $x - 3(7 - 2x) = 7$. This simplifies to $x - 21 + 6x = 7$, or $7x = 28$, so $x = 4$. Then $y = 7 - 2(4) = -1$. The solution is $(4, -1)$.
• To solve the system $a + 3b = 9$ and $2a - b = 4$, solve the second equation for $b$ to get $b = 2a - 4$. Substitute into the first equation: $a + 3(2a - 4) = 9$. This gives $a + 6a - 12 = 9$, so $7a = 21$ and $a=3$. Then $b = 2(3) - 4 = 2$. The solution is $(3, 2).

Property

To Solve a System by Elimination.

1. Write each equation in the form $Ax + By = C$.
2. Decide which variable to eliminate. Multiply each equation by an appropriate constant so that the coefficients of that variable are opposites.
3. Add the equations from Step 2 and solve for the remaining variable.
4. Substitute the value found in Step 3 into one of the original equations and solve for the other variable.

Examples

• To solve the system $2x + 5y = 12$ and $3x - 5y = 3$, the $y$-coefficients are opposites. Add the equations: $(2x+5y) + (3x-5y) = 12+3$, which gives $5x = 15$, so $x=3$. Substitute into the first equation: $2(3) + 5y = 12$, so $5y=6$ and $y=1.2$. The solution is $(3, 1.2)$.
• To solve the system $x + 2y = 8$ and $3x + 4y = 20$, multiply the first equation by $-2$ to get $-2x - 4y = -16$. Add this to the second equation: $(3x+4y) + (-2x-4y) = 20-16$, giving $x = 4$. Substitute into the original first equation: $4 + 2y = 8$, so $2y=4$ and $y=2$. The solution is $(4, 2)$.
• To solve the system $2x + 3y = 7$ and $5x - 2y = 8$, multiply the first equation by $2$ and the second by $3$. This gives $4x + 6y = 14$ and $15x - 6y = 24$. Add them: $19x = 38$, so $x=2$. Substitute into the first original equation: $2(2) + 3y = 7$, so $4+3y=7$, $3y=3$, and $y=1$. The solution is $(2, 1)$.

Property

When we add a multiple of one equation to the other we are making a linear combination of the equations. The method of elimination is also called the method of linear combinations.

Examples

• For the system $2x+y=8$ and $x-y=1$, you can add them to form the linear combination $(2x+y)+(x-y) = 8+1$, which simplifies to the new equation $3x=9$.
• For the system $x+y=5$ and $3x+2y=12$, multiply the first equation by $-2$ to get $-2x-2y=-10$. Then form the linear combination $(-2x-2y)+(3x+2y) = -10+12$, which simplifies to $x=2$.
• For the system $3x+4y=10$ and $2x+3y=7$, multiply the first by $2$ and the second by $-3$. The new equations are $6x+8y=20$ and $-6x-9y=-21$. Add them to get the linear combination $-y=-1$.

Explanation

A linear combination is the process of adding scaled versions of equations together. This is the core engine of the elimination method, allowing us to create a new, simpler equation where one variable has been removed.

Property

When Using Elimination to Solve a System.

1. If combining the two equations results in an equation of the form

then the system is inconsistent.

2. If combining the two equations results in an equation of the form

then the system is dependent.