Question

solve for x. \\(\frac{x}{4} = \frac{2}{x+7}\\)\
\\(\bigcirc\\) x = -1, x = 8\
\\(\bigcirc\\) x = 1, x = -8\
\\(\bigcirc\\) x = 2, x = -4\
\\(\bigcirc\\) x = -2, x = 4

Explanation:

Step1: Cross - multiply the equation

Given the equation \(\frac{x}{4}=\frac{2}{x + 7}\), cross - multiplying (which is based on the property that if \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\)) gives us \(x(x + 7)=4\times2\).

Step2: Expand and simplify the equation

Expanding the left - hand side: \(x^{2}+7x = 8\).
Rearranging the equation to form a quadratic equation in standard form \(ax^{2}+bx + c = 0\), we get \(x^{2}+7x-8 = 0\).

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(-8\) and add up to \(7\). The numbers are \(8\) and \(-1\) because \(8\times(-1)=-8\) and \(8+( - 1)=7\).
So, the factored form of the quadratic equation \(x^{2}+7x - 8=0\) is \((x + 8)(x - 1)=0\).

Step4: Solve for x

Using the zero - product property (if \(ab = 0\), then either \(a = 0\) or \(b = 0\)), we set each factor equal to zero:

• If \(x+8 = 0\), then \(x=-8\).
• If \(x - 1=0\), then \(x = 1\).

Answer:

B. \(x = 1,x=-8\)