Solving Equations

How to Solve an Equation

Steps for Solving an Equation

To find all values of \(x\) that satisfy an equation (e.g., \(f(x) = g(x)\)), complete the following steps:

1. Rewrite equation as \(f(x) - g(x) = 0\), if needed.

2. Simplify and factor the left-hand side.

3. Set each factor equal to zero and solve for \(x\).

Example 1

Solving a quadratic equation

Find all values of \(x\) such that \(x^2 - 4x - 12 = 0\).

Step 1:   Factor   \(x^2 - 4x - 12\).

Recall from Factoring, Example 8 that

\[x^2 - 4x - 12 = (x+2)(x-6)\]

Step 2:   Set each factor equal to zero and solve for   \(x\).

\[\begin{align*} x + 2 &= 0 ~~~~\Rightarrow~~~~ x = -2\\ x - 6 &= 0 ~~~~\Rightarrow~~~~ x = 6 \end{align*}\]

Therefore, \(x=-2\) and \(x=6\) satisfy \(x^2 - 4x - 12 = 0\).

Check Our Work.

We can check our work by plugging in \(x=-2\) and \(x=6\) into the given polynomial and making sure it evaluates to zero.

\[\begin{align*} (-2)^2 - 4(-2) - 12 &= 4 + 8 - 12 = 0 \\ 6^2 - 4(6) - 12 &= 36 - 24 - 12 = 0 \end{align*}\]

Video Resource

Solving Quadratic Equations by Factoring (Links to an external site)
Several more examples of solving quadratic equations.

Example 2

Solving a rational equation

Find all values of \(p\) such that \(\dfrac{3p}{180-6p} = 1\).

Step 1:   Multiply both sides of   \(\frac{3p}{180-6p} = 1\)   by the denominator,   \(180-6p\).

\[3p = 180 - 6p\]

Step 2:   Subtract   \(180 - 6p\)   from both sides and simplify.

\[\begin{align*} 3p - (180-6p) &= 0 \\ 3p - 180 +6p &= 0 && \hbox{distribute the minus sign}\\ 9p - 180 &= 0 && \hbox{combine like terms} \end{align*}\]

Step 3:   Solve for   \(p\).

\[\begin{align*} 9p &= 180 && \text{add $180$ to both sides} \\ p &= 20 && \text{divide both sides by 9} \end{align*}\]

Therefore, \(p=20\) is the only value that satisfies \(\frac{3p}{180-6p} = 1\).

Check Our Work.

We can check our work by plugging in \(p=20\) into \(\dfrac{3p}{180-6p}\) and making sure it evaluates to one.

\[\dfrac{3(20)}{180-6(20)} = \frac{60}{180-120} = \frac{60}{60} = 1\]

Example 3

Points of intersection

Find all points of intersection of \(f(x) = 6x^2 - 4x\) and \(g(x) = 2 - 5x\).

Step 1:   Set   \(f(x) = g(x)\).

Points of intersection can be found by setting the two curves equal to each other and solving for \(x\).

\[6x^2 - 4x = 2 - 5x\]

Step 2:   Subtract   \(2-5x\)   from both sides of the equation in Step 1 and simplify.

\[\begin{align*} 6x^2 - 4x - (2 - 5x) &= 0 && \hbox{note the parentheses around $2-5x$}\\ 6x^2 - 4x - 2 + 5x &= 0 && \hbox{distribute the minus sign}\\ 6x^2 + x - 2 &= 0 && \hbox{combine like terms} \end{align*}\]

Step 3:   Use the AC grouping method to factor   \(6x^2 + x - 2\).

Find two integers that multiply to \(6(-2) = -12\) and sum to \(1\).

Product equals \(-12\)	Sum equals \(1\)?
\(-1 \times 12\)	NO
\(-2 \times 6\)	NO
\(-3 \times 4\)	YES

Therefore,

\[\begin{align*} 6x^2 + x - 2 &= 6x^2 - 3x + 4x - 2 \\ &= 3x(2x-1) + 2(2x-1) \\ &= (3x+2)(2x-1) \end{align*}\]

Step 4:   Set each factor equal to zero and solve for   \(x\).

\[\begin{align*} 3x + 2 &= 0 ~~~~\Rightarrow~~~~ 3x = -2 ~~~~\Rightarrow~~~~ x = -2/3\\ 2x - 1 &= 0 ~~~~\Rightarrow~~~~ 2x = 1 \hspace{25pt}\Rightarrow~~~~ x = 1/2 \end{align*}\]

Therefore, the only points of intersection of \(f(x) = 6x^2 - 4x\) and \(g(x) = 2 - 5x\) occur when \(x=-2/3\) and \(x=1/2\).

Check Our Work.

We can check our work by making sure \(f(x) = g(x)\) when we plug in \(x=-2/3\) and when we plug int \(x=1/2\).

\(x\)	\(f(x) = 6x^2 - 4x\)	\(g(x) = 2 - 5x\)
\(-2/3\)	\(6(-2/3)^2 - 4(-2/3) = 16/3 \)	\(2 - 5(-2/3) = 16/3\)
\(1/2\)	\(6(1/2)^2 - 4(1/2) = -1/2\)	\(2 - 5(1/2) = -1/2\)