 # 3-2 Solving Linear Systems Algebraically Objective: CA 2.0: Students solve system of linear equations in two variables algebraically.

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3-2 Solving Linear Systems Algebraically Objective: CA 2.0: Students solve system of linear equations in two variables algebraically.

Substitution Method 1. Solve one equation for one of its variables 2. Substitute the expression from step 1 into the other equation and solve for the other variable 3. Substitute the value from step 2 into the revised equation from step 1 and solve

Example 1: Solve the linear system of equation using the substitution method.

Step 1: Solve one equation for one of its variables.

Step 2: Substitute the expression from step 1 into the other equation and solve.

Step 3: Substitute the value from step 2 into the revised equation from step 1 and solve. The solution is (-8, 5)

Which equation should you choose in step 1? In general, you should solve for a variable whose coefficient is 1 or -1

The Linear Combination Method Step 1: Multiply one or both of the equations by a constant to obtain coefficients that differ only in sign for one variable.

Step 2: Add the revised equations from step 1. Combining like terms will eliminate one variable. Solve for the remaining variable. Step 3: Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable.

Example 2 Solve the linear system using the Linear Combination (Elimination) Method.

Step 1: Multiply one or both of the equations by a constant to obtain coefficients that differ only in sign for one variable Multiply everything by -2 Leave alone

Step 2: Add the revised equations. After step 1 we now have… -4x + 8y = -26 4x – 5y = 8 Solve for y

Step 3 Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable.

Check your solution The solution checks

Example 3: Linear Combination: Multiply both Equations Solve the linear system using the Linear Combination method.

Step 1) Multiply one or both equations by a constant to obtain coefficients that differ only in sign for one variable.

Step 2) Add the revised equations

Step 3) Substitute the value obtained in Step 2 into either original equation Solution (2, 3)

Example 4 Linear Systems with many or no solutions Solve the linear system.

Use the substitution method Step 1) Step 2) 6 = 7 Because 6 is not equal to 7, there are no solutions.

Two lines that do not intersect are parallel.

Solve the Linear System Solve using the linear combination method Step 1)

Step 2) Add revised equations Because the equation 0 = 0 is always true, there are infinitely many solutions.

Linear Equations that have infinitely many solutions are equivalent equations for the same line.

Home work page 153 12 – 20 even, 24 – 34 even, 38 – 52 even.

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