Question

solve for u.
\\(\frac{4}{u + 10} = \frac{u}{3u + 2}\\)
there may be 1 or 2 solutions.
\\(u = \square\\) or \\(u = \square\\)

Explanation:

Step1: Cross - multiply the fractions

Given the equation \(\frac{4}{u + 10}=\frac{u}{3u+2}\), cross - multiplying (which is valid when \(u+10
eq0\) and \(3u + 2
eq0\), i.e., \(u
eq - 10\) and \(u
eq-\frac{2}{3}\)) gives us \(4\times(3u + 2)=u\times(u + 10)\).

Step2: Expand both sides

Expanding the left - hand side: \(4\times(3u+2)=12u + 8\).
Expanding the right - hand side: \(u\times(u + 10)=u^{2}+10u\).
So our equation becomes \(12u + 8=u^{2}+10u\).

Step3: Rearrange into standard quadratic form

Subtract \(12u\) and \(8\) from both sides to get \(u^{2}+10u-12u - 8 = 0\).
Simplify the like terms: \(u^{2}-2u - 8=0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(-8\) and add up to \(-2\). The numbers are \(-4\) and \(2\).
So, \(u^{2}-2u - 8=(u - 4)(u+2)=0\).

Step5: Solve for u

Set each factor equal to zero:

• If \(u - 4=0\), then \(u = 4\).
• If \(u+2=0\), then \(u=-2\).

We check these solutions in the original equation:

• For \(u = 4\): \(\frac{4}{4 + 10}=\frac{4}{14}=\frac{2}{7}\) and \(\frac{4}{3\times4+2}=\frac{4}{14}=\frac{2}{7}\), so \(u = 4\) is valid.
• For \(u=-2\): \(\frac{4}{-2 + 10}=\frac{4}{8}=\frac{1}{2}\) and \(\frac{-2}{3\times(-2)+2}=\frac{-2}{-4}=\frac{1}{2}\), so \(u=-2\) is valid.

Answer:

\(u = 4\) or \(u=-2\)