Rearranging Formulas: past paper questions & worked solutions

Multiply by the denominator, which is nonzero in the original expression.

Collect both x-terms together and factor out x.

Divide by 2y+3. The inverse formula requires y not equal to -3/2; the original requires x not equal to 2.

Multiply by x^2+6, which is positive for real x.

Collect both terms involving x^2 on one side.

Factor out x^2 and divide by a+1, assuming a is not -1.

\[x=\pm\sqrt{\frac{3w-6a}{a+1}}\quad(a\ne-1).\] For real solutions the radicand must be nonnegative. If \(a=-1\), the original equation instead requires \(w=-2\); then every real \(x\) satisfies it.

Taking square roots gives both signs; handle separately the value excluded by division.

$$p^2 = \\frac{7w + y}{cw + k}$$

**Step 2: Multiply both sides by the denominator (*c**w*+*k*).** *p*²(*c**w*+*k*) = 7*w* + *y*

*p*²*c**w* + *p*²*k* = 7*w* + *y*

**Step 4: Collect all terms involving *w* on one side.** *p*²*c**w* − 7*w* = *y* − *p*²*k*

**Step 5: Factor out *w*.** *w*(*p*²*c*−7) = *y* − *p*²*k*

**Step 6: Divide to isolate *w*.** $$w = \\frac{y - p^2k}{p^2c - 7}$$

Make *k* the subject of $\\displaystyle p = \\dfrac{8k^2+5}{7-3k^2}$.

Multiply both sides by (7−3*k*²): *p*(7−3*k*²) = 8*k*² + 5

Expand: 7*p* − 3*p**k*² = 8*k*² + 5

Collect all *k*² terms on one side: 7*p* − 5 = 8*k*² + 3*p**k*²

Factorise and solve for *k*: $$7p - 5 = k^2(3p+8) \\implies k^2 = \\frac{7p-5}{3p+8}$$ $$k = \\pm\\sqrt{\\dfrac{7p-5}{3p+8}}$$

Include both ± roots unless the context specifies the sign of *k*.

Square both sides: \[ y^2=\frac{x+1}{x-4} \]

(b) \(g^0=1\).

So \(n+2=10\), hence \(n=8\).

Multiply by 15: \[ 3(3x+2)-5(2x+1)=15x \]

The critical values are \(y=-4\) and \(y=\frac83\). Since the quadratic is positive outside the roots, \[ y<-4 \quad \text{or} \quad y>\frac83 \]

Multiply by 20: \[ 4(9a-7)-5(3a-7)=91 \]