Linear Equations by Example -- Part 1.1 --
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If x = 6, then |
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Divide by 3 |
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| 5(x – 2) + 3 = 2(3x – 1) | Original equation |
If x = -5, then |
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| 5x – 10 + 3 = 6x – 2 | Distributive
Property a(b – c) = ab – ac |
5(-5 – 2) + 3 = 2(3(-5) – 1) |
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5x – 7 = 6x – 2 -6x + 7 -6x + 7 -x + 0 = 0 + 5 |
Combine
Like Terms Subtract 6x , Add 7 to both sides |
5(-7) + 3 = 2(-15 – 1) |
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-x = 5 x = -5 |
Multiply
both sides by -1 -x = -1 × x |
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x + 4 = x + 8 |
Original equation |
No solution |
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x + 4 = x + 8 |
Subtract
x
from both sides But 4 ¹ 8 |
There is no number plus 4 can equal the same number plus 8 | ||||
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| Not all linear equations have solutions. Can you think of an linear equation that would have more than one solutions. | ||||||
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Original
equation
Multiply both sides by the LCD, 15, to clear fractions. Divide by 11 LCD = 3
× 5 × 1 = 15 |
Rewrite |
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[Solution] |
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Factor denominator. Note: x can not be 5 or -5. You can't have zero in the denominator. | |||||
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Multiply both sides by the LCD to clear fractions. | |||||
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LCD = (x + 5)(x – 5) | |||||
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4(x – 5) + 2(x + 5) = 32 |
Check |
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4x – 20 + 2x + 10 = 32 |
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6x = 42 |
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[Solution] |
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Send comments, suggestions or report errors to
joe mcdonald
Last Update 03.15.2006