smc-4402-40-Multiple fractions

v1 Functions, 2ADV F1 2004 HSC 1c

Solve `(x+4)/5-(x-2)/6 = 4`. (2 marks)

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`38`

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`(x+4)/5-(x-2)/6`	`= 4`
`30((x+4)/5)-30((x-2)/6)`	`= 30 xx 4`
`6x+24-5x+10`	`= 120`
`x+34`	`= 120`
`:. x`	`= 38`

Algebraic Fractions, SMB-050

Solve the equation `(2p+2)/3+1 = (p-5)/5`, leaving your answer as a fraction. (3 marks)

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`p=-40/7`

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`underbrace{(2p+2)/3+1}_text(multiply × 15)`	`=underbrace{(p-5)/5}_text(multiply × 15)`
`5(2p+2)+15`	`= 3(p-5)`
`10p+10+15`	`=3p-15`
`7p`	`= -40`
`p`	`=-40/7`

Algebraic Fractions, SMB-049

Solve the equation `(x-1)/2+(2x+3)/3 = 2`, leaving your answer as a fraction. (3 marks)

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`x=9/7`

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`underbrace{(x-1)/2+(2x+3)/3}_text(multiply × 6)`	`=2xx6`
`3(x-1)+2(2x+3)`	`= 12`
`3x-3+4x+6`	`=12`
`7x`	`= 9`
`x`	`=9/7`

Algebraic Fractions, SMB-048

Solve `(2x+1)/3-(x+1)/8=1` for `x`. (3 marks)

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`x=19/13`

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`(2x+1)/3-(x+1)/8`	`=1`
`(24(2x+1))/3-(24(x+1))/8`	`=24`
`8(2x+1)-3(x+1)`	`=24`
`16x+8-3x-3`	`=24`
`13x`	`=19`
`y`	`=19/13`

Algebraic Fractions, SMB-047

Solve the equation `(3a)/7 = (2a + 1)/2-3`, leaving your answer as a fraction. (3 marks)

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`a=35/8`

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`underbrace{(3a)/7}_text(multiply × 14)`	`=underbrace{(2a + 1)/2-3}_text(multiply × 14)`
`2xx3a`	`= 7xx(2a+1)-14xx3`
`6a`	`=14a+7-42`
`8a`	`= 35`
`a`	`=35/8`

Algebraic Fractions, SMB-046

Solve `(3y-1)/4-(y+1)/3=2` for `y`. (3 marks)

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`y=31/5`

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`(3y-1)/4-(y+1)/3`	`=2`
`(12(3y-1))/4-(12(y+1))/3`	`=24`
`3(3y-1)-4(y+1)`	`=24`
`9y-3-4y-4`	`=24`
`5y`	`=31`
`y`	`=31/5`

Algebraic Fractions, SMB-045

Solve `(2a-5)/3-(a+7)/5=3` for `a`. (3 marks)

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`a=13`

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`(2a-5)/3-(a+7)/5`	`=3`
`(5(2a-5))/15-(3(a+7))/15`	`=3`
`5(2a-5)-3(a+7)`	`=45`
`10a-25-3a-21`	`=45`
`7a`	`=91`
`a`	`=13`

Algebra, STD2 A1 2018 HSC 28b

Solve the equation `(2x)/5 + 1 = (3x + 1)/2`, leaving your answer as a fraction. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

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`5/11`

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♦ Mean mark 35%.

`underbrace{(2x)/5 + 1}_text(multiply x10)`	`=underbrace{(3x + 1)/2}_text(multiply x10)`
`4x + 10`	`= 15x + 5`
`11x`	`= 5`
`x`	`= 5/11`

Algebra, STD2 A1 2015 HSC 24 MC

Consider the equation `(2x)/3-4 = (5x)/2 + 1`.

Which of the following would be a correct step in solving this equation?

`(2x)/3-3 = (5x)/2`
`(2x)/3 = (5x)/2 + 5`
`2x-4 = (15x)/2 + 3`
`(4x)/6-8 = 5x + 2`

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`B`

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`(2x)/3-4`	`= (5x)/2 + 1`
`(2x)/3`	`= (5x)/2 + 5`

`=>B`

Functions, 2ADV F1 2004 HSC 1c

Solve `(x-5)/3-(x+1)/4 = 5`. (2 marks)

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`83`

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`(x-5)/3-(x+1)/4`	`= 5`
`12((x-5)/3)-12((x+1)/4)`	`= 12 xx 5`
`4x-20-3x-3`	`= 60`
`x-23`	`= 60`
`:. x`	`= 83`

Algebra, STD2 A1 2013 HSC 29a

Sarah tried to solve this equation and made a mistake in Line 2.

`(W+4)/3-(2W-1)/5`	`=1`	`text(... Line 1)`
`5W+ 20-6W-3`	`=15`	`text(... Line 2)`
`17-W`	`=15`	`text(... Line 3)`
`W`	`=2`	`text(... Line 4)`

Copy the equation in Line 1 and continue your solution to solve this equation for `W`.

Show all lines of working. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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`(W+4)/3-(2W-1)/5`	`=1`	`text(… Line 1)`
`5W+ 20-6W+ 3`	`=15`	`text(… Line 2)`
`23-W`	`=15`	`text(… Line 3)`
`W`	`=8`	`text(… Line 4)`

Show Worked Solution

♦♦ Mean mark 27%
STRATEGY: The RHS of the equation increases from 1 to 15 (from Line 1 to Line 2), indicating both sides must have been multiplied by 15.

`(W+4)/3-(2W-1)/5`	`=1`	`text(… Line 1)`
`5W+ 20-6W+3`	`=15`	`text(… Line 2)`
`23-W`	`=15`	`text(… Line 2)`
`W`	`=8`	`text(… Line 4)`