Algebra Examples

Solve by Addition/Elimination 6a+6b=12 6a-5b=12
Step 1
Multiply each equation by the value that makes the coefficients of opposite.
Step 2
Simplify.
Tap for more steps...
Step 2.1
Simplify the left side.
Tap for more steps...
Step 2.1.1
Simplify .
Tap for more steps...
Step 2.1.1.1
Apply the distributive property.
Step 2.1.1.2
Multiply.
Tap for more steps...
Step 2.1.1.2.1
Multiply by .
Step 2.1.1.2.2
Multiply by .
Step 2.2
Simplify the right side.
Tap for more steps...
Step 2.2.1
Multiply by .
Step 3
Add the two equations together to eliminate from the system.
Step 4
Divide each term in by and simplify.
Tap for more steps...
Step 4.1
Divide each term in by .
Step 4.2
Simplify the left side.
Tap for more steps...
Step 4.2.1
Cancel the common factor of .
Tap for more steps...
Step 4.2.1.1
Cancel the common factor.
Step 4.2.1.2
Divide by .
Step 4.3
Simplify the right side.
Tap for more steps...
Step 4.3.1
Divide by .
Step 5
Substitute the value found for into one of the original equations, then solve for .
Tap for more steps...
Step 5.1
Substitute the value found for into one of the original equations to solve for .
Step 5.2
Simplify .
Tap for more steps...
Step 5.2.1
Multiply by .
Step 5.2.2
Add and .
Step 5.3
Divide each term in by and simplify.
Tap for more steps...
Step 5.3.1
Divide each term in by .
Step 5.3.2
Simplify the left side.
Tap for more steps...
Step 5.3.2.1
Cancel the common factor of .
Tap for more steps...
Step 5.3.2.1.1
Cancel the common factor.
Step 5.3.2.1.2
Divide by .
Step 5.3.3
Simplify the right side.
Tap for more steps...
Step 5.3.3.1
Divide by .
Step 6
This is the final solution to the independent system of equations.