Equations with fractions or decimals

When an equation contains fractions or decimals, first change it into an equivalent equation with whole-number coefficients. Then solve by elimination or substitution as usual.

Why it is allowed: multiplying every term by the same non-zero number keeps the equation equivalent. Multiplying by the lowest common multiple (LCM) of the denominators clears all the fractions at once; multiplying by \(10\) or \(100\) clears decimals.

Method

1. For each equation separately, find the LCM of its denominators, or the power of \(10\) that clears its decimals.
2. Multiply every term of that equation by it, including any term without a fraction.
3. Solve the new pair by elimination or substitution.
4. Check the answer in the original equations with the fractions.

Example

Solve \(\dfrac{x}{5}+\dfrac{y}{7}=2\) \((1)\) and \(\dfrac{x+3}{2}=y-3\) \((2)\).

\(\begin{aligned} (1)\times35:\quad 7x+5y&=70\quad(3) \\ (2)\times2:\quad x+3&=2y-6 \\ x-2y&=-9\quad(4) \end{aligned}\)

From \((4)\): \(x=2y-9\). Substitute into \((3)\):

\(\begin{aligned} 7(2y-9)+5y&=70 \\ 19y-63&=70 \\ y&=7 \end{aligned}\)

Then \(x=2(7)-9=5\). Check in \((1)\): \(\dfrac{5}{5}+\dfrac{7}{7}=1+1=2\). Check in \((2)\): \(\dfrac{5+3}{2}=4\) and \(7-3=4\).

For decimals, \(0.6x+0.5y=2.8\) becomes \(6x+5y=28\) after multiplying by \(10\).

When the unknown is in a denominator (here \(y\ne-1\)), cross-multiply first. Solve \(\dfrac{x-1}{y+1}=5\) and \(x-y=10\).

\(\begin{aligned} x-1&=5(y+1) \\ x-5y&=6 \end{aligned}\)

Subtracting this from \(x-y=10\) gives \(4y=4\), so \(y=1\) and \(x=11\). Check: \(\dfrac{11-1}{1+1}=\dfrac{10}{2}=5\) and \(11-1=10\).

Check: multiply the whole-number terms too. In \((1)\times35\), the right-hand side becomes \(70\), not \(2\). A common error is to multiply only the fractions.