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Math · College algebra · Worked example

Solve a system of equations by elimination

Solve the system 2x + 3y = 12 and x − y = 1.

2⁢x+3⁢y=12,x−y=1

Make the y-coefficients opposites

Multiply every term of x − y = 1 by 3, so that its y-term, −3y, is the opposite of 3y in the first equation.

3⁢x−3⁢y=3

Add the equations

(2x + 3y) + (3x − 3y) = 12 + 3. The y-terms cancel, leaving 5x = 15, so x = 3.

5⁢x=15⟹x=3

Find y

Substitute x = 3 into x − y = 1: 3 − y = 1, so y = 2.

3−y=1⟹y=2

Check in both original equations

2(3) + 3(2) = 12 and 3 − 2 = 1, so (3, 2) satisfies both equations.

2⁢(3)+3⁢(2)=123−2=1

Result

x = 3 and y = 2: the lines cross at (3, 2).

Your turn

Solve 4x − y = 7 and 2x + 3y = 21.

Show the answer and explanation

x = 3 and y = 5.

From the first equation, y = 4x − 7. Substituting into the second gives 2x + 3(4x − 7) = 21, so 14x = 42 and x = 3; then y = 4(3) − 7 = 5. Check: 4(3) − 5 = 7 and 2(3) + 3(5) = 21.

4⁢(3)−5=72⁢(3)+3⁢(5)=21

Keep exploring

In Matrices & linear systems, change the second row to 2, 3, 5, the equation 2x + 3y = 5. The coefficients now match the first row but the constants differ, and the studio reports no solution: the lines are parallel.

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