Question

Determine all values of x that make the expression true. work in the space provided.

Answer 1

The equation is given as shown below:

$\frac{2}{x-3}=\frac{x+4}{x}$

Step 1

Apply fraction cross multiply, defined as:

$\begin{gathered} If \\ \frac{a}{b}=\frac{c}{d} \\ \text{then} \\ a\cdot d=b\cdot c \end{gathered}$

Thus, we have:

$\Rightarrow2x=(x-3)(x+4)$

Step 2

Expand the right-hand side of the equation using the FOIL method given to be:

$(a+b)(c+d)=ac+ad+bc+bd$

Thus, we have:

$(x-3)(x+4)=x^2+4x-3x-12=x^2+x-12$

Therefore, the expression becomes:

$2x=x^2+x-12$

Step 3

Write out the equation in the standard form of a quadratic equation given to be:

$ax^2+bx+c=0$

Hence, we have:

$\begin{gathered} x^2+x-2x-12=0 \\ x^2-x-12=0 \end{gathered}$

Step 4

Solve using the quadratic formula given to be:

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Given

$\begin{gathered} a=1 \\ b=-1 \\ c=-12 \end{gathered}$

Hence, we have:

$\begin{gathered} x=\frac{-(-1)\pm\sqrt[]{(-1)^2-(4\times1\times\lbrack-12\rbrack)}}{2\times1}=\frac{1\pm\sqrt[]{1+48}}{2}=\frac{1\pm\sqrt[]{49}}{2} \\ x=\frac{1\pm7}{2} \end{gathered}$

Therefore, the values of x can be:

$\begin{gathered} x=\frac{1+7}{2}=\frac{8}{2}=4 \\ or \\ x=\frac{1-7}{2}=\frac{-6}{2}=-3 \end{gathered}$

Therefore, the possible solutions to the equation are:

$x=4,x=-3$

Step 5

Check for undefined points. This we can do by equating the denominator of the equation to zero.

Therefore, the undefined points are at:

$\begin{gathered} x-3=0 \\ \therefore \\ x=3 \end{gathered}$

or

$x=0$

ANSWER

Combining the solution and the undefined points, given that none of the solutions are at the undefined points, the solution to the expression is given to be:

$x=4,x=-3$