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Algebra, STD2 A1 2019 HSC 11 MC (Adapted)

Which of the following correctly expresses \(y\) as the subject of the formula  \(5x-2y-9=0\)?

  1.  \(y=\dfrac{5}{2}x-9\)
  2.  \(y=\dfrac{5}{2}x+9\)
  3.  \(y=\dfrac{5x+9}{2}\)
  4.  \(y=\dfrac{5x-9}{2}\)
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\(D\)

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♦ Mean mark 50%.

\(5x-2y-9\) \(=0\)
\(2y\) \(=5x-9\)
\(\therefore\ y\) \(=\dfrac{5x-9}{2}\)

 
\(\Rightarrow D\)

Algebra, STD1 A1 2019 HSC 34 (Adapted)

Given the formula  \(D=\dfrac{B(x+1)}{18}\), calculate the value of  \(x\)  when  \(D=90\)  and  \(B=400\).   (3 marks)

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\(3.05\)

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\(\text{Make}\ x\ \text{the subject:}\)

\(D\) \(=\dfrac{B(x+1)}{18}\)
\(18D\) \(=B(x+1)\)
\(x+1\) \(=\dfrac{18D}{B}\)
\(x\) \(=\dfrac{18D}{B}-1\)
\(\text{When }\) \(D=90, B=400\)
\(\therefore\ x\) \(=\dfrac{18\times 90}{400}-1=3.05\)

Algebra, STD2 A1 2022 HSC 14 MC (Adapted)

Which of the following correctly expresses \(x\) as the subject of  \(y=\dfrac{mx-c}{3}\) ?

  1. \(x=\dfrac{3y}{m}+c\)
  2. \(x=\dfrac{y}{3m}+c\)
  3. \(x=\dfrac{y+c}{3m}\)
  4. \(x=\dfrac{3y+c}{m}\)
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\(D\)

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\(y\) \(=\dfrac{mx-c}{3}\)
\(3y\) \(=mx-c\)
\(mx\) \(=3y+c\)
\(\therefore\ x\) \(=\dfrac{3y+c}{m}\)

 
\(\Rightarrow D\)