Algebra, NAPX-H4-CA18

Question

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Worked Solution

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Variant 0

DifficultyLevel

668

Question

Rearrange the equation  6y\ 6\large y + 7 = x\large x so that y\large y is the subject.

Which of these correctly gives y\large y as the subject?

Worked Solution
6y6\large y + 7 = x\large x
6y6\large y = x\large x −- 7
y\large y = x−76\dfrac{x-7}{6}

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Rearrange the equation $\ 6\large y$ + 7 = $\large x$ so that $\large y$ is the subject. Which of these correctly gives $\large y$ as the subject?
workedSolution
| | | | ------------: | ---------- | | $6\large y$ + 7 | \= $\large x$ | | $6\large y$ | \= $\large x$ $-$ 7 | | $\large y$ | \= $\dfrac{x-7}{6}$ |
correctAnswer
$\large y$ = $\dfrac{x-7}{6}$

Answers

Is Correct?Answer
✓

y\large y = x−76\dfrac{x-7}{6}

x

y\large y = x+76\dfrac{x+7}{6}

x

y\large y = 6x6\large x −- 7

x

y\large y = 6x\large x + 7

Variant 1

DifficultyLevel

668

Question

Rearrange the equation  3m\ 3\large m + 4 = n\large n so that m\large m is the subject.

Which of these correctly gives m\large m as the subject?

Worked Solution
3m3\large m + 4 = n\large n
3m3\large m = n\large n −- 4
m\large m = n−43\dfrac{n-4}{3}

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Rearrange the equation $\ 3\large m$ + 4 = $\large n$ so that $\large m$ is the subject. Which of these correctly gives $\large m$ as the subject?
workedSolution
| | | | ------------: | ---------- | | $3\large m$ + 4 | \= $\large n$ | | $3\large m$ | \= $\large n$ $-$ 4 | | $\large m$ | \= $\dfrac{n-4}{3}$ |
correctAnswer
$\large m$ = $\dfrac{n-4}{3}$

Answers

Is Correct?Answer
x

m\large m = n−34\dfrac{n-3}{4}

✓

m\large m = n−43\dfrac{n-4}{3}

x

m\large m = 3n\large n −- 4

x

m\large m = 3n\large n + 4

Variant 2

DifficultyLevel

668

Question

Rearrange the equation  4y\ 4\large y −- 3 = x\large x so that y\large y is the subject.

Which of these correctly gives y\large y as the subject?

Worked Solution
4y4\large y −- 3 = x\large x
4y4\large y = x\large x + 3
y\large y = x+34\dfrac{x + 3}{4}

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Rearrange the equation $\ 4\large y$ $-$ 3 = $\large x$ so that $\large y$ is the subject. Which of these correctly gives $\large y$ as the subject?
workedSolution
| | | | ------------: | ---------- | | $4\large y$ $-$ 3 | \= $\large x$ | | $4\large y$ | \= $\large x$ + 3 | | $\large y$ | \= $\dfrac{x + 3}{4}$ |
correctAnswer
$\large y$ = $\dfrac{x+3}{4}$

Answers

Is Correct?Answer
x

y\large y = x−34\dfrac{x-3}{4}

x

y\large y = 4x4\large x −- 3

✓

y\large y = x+34\dfrac{x+3}{4}

x

y\large y = 4x\large x + 3

Variant 3

DifficultyLevel

670

Question

Rearrange the equation  5y\ 5\large y −- 7 = 2x2\large x so that y\large y is the subject.

Which of these correctly gives y\large y as the subject?

Worked Solution
5y5\large y −- 7 = 2x2\large x
5y5\large y = 2x2\large x + 7
y\large y = 2x+75\dfrac{2x+7}{5}

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Rearrange the equation $\ 5\large y$ $-$ 7 = $2\large x$ so that $\large y$ is the subject. Which of these correctly gives $\large y$ as the subject?
workedSolution
| | | | ------------: | ---------- | | $5\large y$ $-$ 7 | \= $2\large x$ | | $5\large y$ | \= $2\large x$ + 7 | | $\large y$ | \= $\dfrac{2x+7}{5}$ |
correctAnswer
$\large y$ \= $\dfrac{2x+7}{5}$

Answers

Is Correct?Answer
x

y\large y = 2x\large x + 7

x

y\large y = 2x\large x −- 7

x

y\large y = 2x−75\dfrac{2x-7}{5}

✓

y\large y = 2x+75\dfrac{2x+7}{5}

Variant 4

DifficultyLevel

670

Question

Rearrange the equation  6y\ 6\large y + x\large x = 5 so that y\large y is the subject.

Which of these correctly gives y\large y as the subject?

Worked Solution
 6y\ 6\large y + x\large x = 5
6y6\large y = 5 −- x\large x
y\large y = 5−x6\dfrac{5-\large x}{6}

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Rearrange the equation $\ 6\large y$ + $\large x$ = 5 so that $\large y$ is the subject. Which of these correctly gives $\large y$ as the subject?
workedSolution
| | | | ------------: | ---------- | | $\ 6\large y$ + $\large x$ | \= 5 | | $6\large y$ | \= 5 $-$ $\large x$ | | $\large y$ | \= $\dfrac{5-\large x}{6}$ |
correctAnswer
$\large y$ \= $\dfrac{5-\large x}{6}$

Answers

Is Correct?Answer
x

y\large y = 5 −- 6x6\large x

x

y\large y = 5 + 6x6\large x

✓

y\large y = 5−x6\dfrac{5-\large x}{6}

x

y\large y = 5+x6\dfrac{5+\large x}{6}

Variant 5

DifficultyLevel

672

Question

Rearrange the equation  6y\ 6\large y −- x\large x −- 2 = 0 so that y\large y is the subject.

Which of these correctly gives y\large y as the subject?

Worked Solution
 6y\ 6\large y −- x\large x −- 2 = 0
6y6\large y = x\large x ++ 2
y\large y = x+26\dfrac{x+2}{6}

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Rearrange the equation $\ 6\large y$ $-$ $\large x$ $-$ 2 = 0 so that $\large y$ is the subject. Which of these correctly gives $\large y$ as the subject?
workedSolution
| | | | ------------: | ---------- | | $\ 6\large y$ $-$ $\large x$ $-$ 2 | \= 0 | | $6\large y$ | \= $\large x$ $+$ 2 | | $\large y$ | \= $\dfrac{x+2}{6}$ |
correctAnswer
$\large y$ \= $\dfrac{x + 2}{6}$

Answers

Is Correct?Answer
x

y\large y = 2 −- 6x6\large x

x

y\large y = 2 + 6x6\large x

x

y\large y = x−26\dfrac{x - 2}{6}

✓

y\large y = x+26\dfrac{x + 2}{6}

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