from the figure alongside find x
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In ∆ ABC, DE || BC
Applying
AD / BD = AE / EC
( 4x - 3 ) / ( 3x - 1 ) = ( 8x - 7 ) ( 5x - 1 )
Cross - Multiplication
( 4x - 3 ) ( 5x - 1 ) = ( 8x - 7 ) ( 3x - 1 )
4x ( 5x - 1 ) - 3 ( 5x - 1 ) = 8x ( 3x - 1 ) - 7 ( 3x - 1 )
20x² - 4x - 15x + 3 = 24x² - 8x - 21x + 7
20x² - 19x + 3 = 24x² - 29x + 7
24x² - 20x² - 29x + 19x + 7 - 3 = 0
4x² - 10x + 4 = 0
2 ( 2x² - 5x + 2 ) = 0
2x² - 5x + 2 = 0/2
2x² - 5x + 2 = 0
Using
2x² - 4x - x + 2 = 0
2x ( x - 2 ) - 1 ( x - 2 ) = 0
( 2x - 1 ) ( x - 2 ) = 0
Using
2x - 1 = 0 and x - 2 = 0
x = 1/2 and x = 2
Applying
AD / BD = AE / EC
( 4x - 3 ) / ( 3x - 1 ) = ( 8x - 7 ) ( 5x - 1 )
Cross - Multiplication
( 4x - 3 ) ( 5x - 1 ) = ( 8x - 7 ) ( 3x - 1 )
4x ( 5x - 1 ) - 3 ( 5x - 1 ) = 8x ( 3x - 1 ) - 7 ( 3x - 1 )
20x² - 4x - 15x + 3 = 24x² - 8x - 21x + 7
20x² - 19x + 3 = 24x² - 29x + 7
24x² - 20x² - 29x + 19x + 7 - 3 = 0
4x² - 10x + 4 = 0
2 ( 2x² - 5x + 2 ) = 0
2x² - 5x + 2 = 0/2
2x² - 5x + 2 = 0
Using
2x² - 4x - x + 2 = 0
2x ( x - 2 ) - 1 ( x - 2 ) = 0
( 2x - 1 ) ( x - 2 ) = 0
Using
2x - 1 = 0 and x - 2 = 0
x = 1/2 and x = 2
22072003:
I already the method, U first check my answer step-by-step, if there is any mistake, then tell me.
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