Algebra, STD1 A1 2019 HSC 34 v1

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

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