• Solve equations with fraction coefficients
  • Solve equations with decimal coefficients

Assignment

Solve Equations with Fraction Coefficients

Fractions can be intimidating, but are easy to deal with if you take the right approach. While everything we’ve gone over still applies, we’re going to look at an optional step that will turn your fractions into integers.

\[\begin{align} \frac{1}{8}x + \frac{1}{2} &= \frac{1}{4} \\ 8\cdot \left(\frac{1}{8}x + \frac{1}{2}\right) &= \left(\frac{1}{4}\right) \cdot 8 \\ x + 4 &= 2 \label{gcf}\\ x &= -2 \end{align}\]

Step \ref{gcf} comes from multiplying by the least common denominator of all the fractions present in the problem. This wipes out all the denominators as each coefficient and constant simplifies to an integer.

Here’s another one.

\[\begin{align} 6 &= \frac{1}{2}x + \frac{2}{5}x - \frac{3}{4}x \\ 20\cdot(6) &= \left(\frac{1}{2}x + \frac{2}{5}x - \frac{3}{4}x\right) \cdot 20 \\ 120 &= 10x + 8x - 15x \\ 120 &= 3x \\ 40 &= x \end{align}\]

Both equations we’ve looked at where at the point where we could only combine like terms. If distribution is possible, you can still apply this method, but make sure you only multiply each term only once.

\[\begin{align} \frac{1}{2}(x - 5) &= \frac{1}{4}(x - 1) \label{two-terms}\\ 4 \cdot \left(\frac{1}{2}(x - 5)\right) &= \left(\frac{1}{4}(x - 1)\right)\cdot4 \\ 2(x-5)&=(x-1) \\ 2x-10 &= x - 1 \\ x &= 9 \end{align}\]

There are only two terms in equation \ref{two-terms}; what’s inside the parentheses doesn’t count. If you choose to distribute, then it would be a different story.

\[\begin{align} \frac{1}{2}(x - 5) &= \frac{1}{4}(x - 1)\\ \frac{1}{2}x - \frac{5}{2} &= \frac{1}{4}x - \frac{1}{4} \\ 4 \cdot \left(\frac{1}{2}x - \frac{5}{2}\right) &= \left(\frac{1}{4}x - \frac{1}{4}\right) \cdot 4 \\ 2x - 10 &= x - 1 \\ x &= 9 \end{align}\]

After distributing, we end up with four terms, and now all four multiplied by $4$. Either approach is fine, just be aware that this is not a necessary step. It will make the arithmetic easier, but done incorrectly in can cause problems.

Solve Equations with Decimal Coefficients

We can apply a similar step to equations with decimals. Here, you want to move the decimal over far enough so everything ends up as integer. So, if you need to move it over once, multiply by $10$. Twice? $100$ Thrice? $1000$.

\[\begin{align} 0.06x + 0.02 &= 0.26x - 1.58 \\ 100 \cdot (0.06x + 0.02) &= (0.26x - 1.58) \cdot 100 \\ 6x + 2 &= 26x - 158 \\ 160 &= 20x \\ 8 &= x \end{align}\]