Grade 8 / Solving Systems of Equations 3 / 6

Simultaneous Equations ③ Elimination Method ② (Matching Coefficients)

Simultaneous Equations ③ Elimination Method ②

When the coefficients of the variable to be eliminated are different, multiply one or both equations by suitable numbers to make the coefficients equal, and then add or subtract.

Multiplying One Equation

Example ①
{ 2x+y=7 …①、x+3y=9 …②
Answer:①×3：6x+3y=21 …③。③−② → 5x=12 → x=12/5。…
Alternative: ②×2：2x+6y=18 …③。③−① → 5y=11 → y=11/5。

Example ② (Integer Solution)
{ 3x+2y=12 …①、2x+y=7 …②
Answer:②×2：4x+2y=14 …③。③−① → x=2. Substitute into ①: y=3. (2,3)

Multiplying Both Equations (Using the LCM)

Example ③
{ 2x+3y=13 …①、3x+2y=12 …②
Answer:①×3：6x+9y=39. ②×2：6x+4y=24. Subtract: 5y=15→y=3. Substitute into ②→x=2. (2,3)

Key Points

Match the coefficients to the LCM of the coefficients of the variable to be eliminated
If multiplying only one equation by a constant is enough, that method is easier
Pay attention to the signs after multiplication (same signs → subtract, different signs → add)

Practice Problems

Solve { 3x+2y=16、x+y=6 using the elimination method.
Solve { 2x+3y=7、3x+2y=8 using the elimination method.

Answers and Explanations

Answer:(x,y)=(4,2)。②×2：2x+2y=12。①−②×2：(3x+2y)−(2x+2y)=16−12→x=4. Substitute into ②: 4+y=6→y=2. Therefore (4,2)。
Answer:(x,y)=(2,1)。①×3−②×2→9y−4y=21−16→5y=5→y=1. Substitute into ①→x=2。

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Simultaneous Equations ② Elimination Method ① (Equal or Opposite Coefficients)

Simultaneous Equations ④ Substitution Method

Simultaneous Equations ③ Elimination Method ② (Matching Coefficients)