(a) If p varies directly as \(r^{2}\) and p = 3.2 when r = 4, find the value of p when r = 6.5. (b) Solve the simultaneous equations : \(\frac{x}{2} + \frac...
Question 1 Report
(a) If p varies directly as \(r^{2}\) and p = 3.2 when r = 4, find the value of p when r = 6.5.
(b) Solve the simultaneous equations :
\(\frac{x}{2} + \frac{y}{4} = 1 ; \frac{x}{3} - \frac{y}{4} = \frac{-1}{6}\)
Answer Details
(a) \(p\propto r^{2}\Rightarrow p=kr^{2}\).
Using \(p=3.2,\ r=4\): \(3.2=k(16)\Rightarrow k=0.2\).
When \(r=6.5\): \(p=0.2(6.5)^{2}=0.2(42.25)=8.45\).
\(p=8.45\)
(b) Clear the fractions.
\(\dfrac{x}{2}+\dfrac{y}{4}=1\) \(\times4\Rightarrow 2x+y=4\) ... (1)
\(\dfrac{x}{3}-\dfrac{y}{4}=-\dfrac{1}{6}\) \(\times12\Rightarrow 4x-3y=-2\) ... (2)
From (1): \(y=4-2x\). Substitute into (2):
\[4x-3(4-2x)=-2\Rightarrow 4x-12+6x=-2\Rightarrow 10x=10\Rightarrow x=1\]Then \(y=4-2(1)=2\).
\(x=1,\ y=2\)
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