Asked by k.
1. Why does the elimination method work when solving a system of equations?
2. Esteban tries to solve the following system. 7x - 4y = -12 x - 2y = 4 His first step is to multiply the second equation by 3. 7x - 4y = -12 3x - 6y = 12 Then adds the equations to eliminate a term. What is Esteban's error?
5. 4x - 2y = -2 3x + 2y = -12
6. 3x + 2y = 4 3x + 6y = -24
8. x - 3y = -4 2x - 6y = 6
All Answers 1
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1) Why elimination works
- Each linear equation is a statement that restricts x and y. If a pair (x,y) satisfies both equations, it also satisfies any equation you get by adding or subtracting a nonzero multiple of one equation to/from the other.
- Algebraically: if A(x,y)=0 and B(x,y)=0, then for any nonzero constant k, A + kB = 0 is true for the same (x,y). Thus forming linear combinations produces new equations that are true exactly for the same solution set. By choosing multipliers so a variable's coefficients become opposites, adding the equations cancels that variable and yields a single-variable equation you can solve. (Do not multiply by 0, since that loses information.)
2) Esteban's error
- He multiplied x - 2y = 4 by 3 correctly to get 3x - 6y = 12, but those y-coefficients (-4 and -6) are not opposites, so adding the two equations does not eliminate y. He picked the wrong multiplier. To eliminate y he should have multiplied the second equation by 2 (giving 2x - 4y = 8) and then subtracted or added appropriately.
Problems
5) 4x - 2y = -2
3x + 2y = -12
Add to eliminate y: 7x = -14, so x = -2.
Substitute: 3(-2) + 2y = -12 → -6 + 2y = -12 → 2y = -6 → y = -3.
Solution: (x,y) = (-2, -3).
6) 3x + 2y = 4
3x + 6y = -24
Subtract the first from the second to eliminate x: 4y = -28 → y = -7.
Substitute: 3x + 2(-7) = 4 → 3x - 14 = 4 → 3x = 18 → x = 6.
Solution: (x,y) = (6, -7).
8) x - 3y = -4
2x - 6y = 6
Multiply the first by 2: 2x - 6y = -8, which contradicts 2x - 6y = 6. The left-hand sides are identical but the right-hand sides differ, so the system is inconsistent.
Conclusion: no solution (the lines are parallel).
- Each linear equation is a statement that restricts x and y. If a pair (x,y) satisfies both equations, it also satisfies any equation you get by adding or subtracting a nonzero multiple of one equation to/from the other.
- Algebraically: if A(x,y)=0 and B(x,y)=0, then for any nonzero constant k, A + kB = 0 is true for the same (x,y). Thus forming linear combinations produces new equations that are true exactly for the same solution set. By choosing multipliers so a variable's coefficients become opposites, adding the equations cancels that variable and yields a single-variable equation you can solve. (Do not multiply by 0, since that loses information.)
2) Esteban's error
- He multiplied x - 2y = 4 by 3 correctly to get 3x - 6y = 12, but those y-coefficients (-4 and -6) are not opposites, so adding the two equations does not eliminate y. He picked the wrong multiplier. To eliminate y he should have multiplied the second equation by 2 (giving 2x - 4y = 8) and then subtracted or added appropriately.
Problems
5) 4x - 2y = -2
3x + 2y = -12
Add to eliminate y: 7x = -14, so x = -2.
Substitute: 3(-2) + 2y = -12 → -6 + 2y = -12 → 2y = -6 → y = -3.
Solution: (x,y) = (-2, -3).
6) 3x + 2y = 4
3x + 6y = -24
Subtract the first from the second to eliminate x: 4y = -28 → y = -7.
Substitute: 3x + 2(-7) = 4 → 3x - 14 = 4 → 3x = 18 → x = 6.
Solution: (x,y) = (6, -7).
8) x - 3y = -4
2x - 6y = 6
Multiply the first by 2: 2x - 6y = -8, which contradicts 2x - 6y = 6. The left-hand sides are identical but the right-hand sides differ, so the system is inconsistent.
Conclusion: no solution (the lines are parallel).
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