Math 2.0

Problem

Problem in words

Problem

In 10 years, the father will be twice as old as his son.

Equivalent equation

y+10=2×(x+10)y=2×(x+10)−10Subtract 10 from both sidesy=2x+20−10Multiple out bracketsy=2x+10Simplify\begin{aligned} y + 10 &= 2 \times (x + 10)\\ y &= 2 \times (x + 10) - 10 && \text{Subtract 10 from both sides}\\ y &= 2x + 20 - 10 && \text{Multiple out brackets}\\ y &= 2x + 10 && \text{Simplify} \end{aligned}

Now there are two related linear equations, each with two unknowns, which enables the production of a linear equation with just one variable, by subtracting one from the other (called the elimination method):

{y=x+22First equationy=2x+10Second equation\begin{cases} y = x + 22 & \text{First equation}\\ y = 2x + 10 & \text{Second equation} \end{cases}

Subtract the first equation from(y−y)=(2x−x)+10−22the second in order to remove y0=x−12Simplify12=xAdd 12 to both sidesx=12Rearrange\begin{aligned} &&&\text{Subtract the first equation from}\\ (y - y) &= (2x - x) +10 - 22 && \text{the second in order to remove } y\\ 0 &= x - 12 && \text{Simplify}\\ 12 &= x && \text{Add 12 to both sides}\\ x &= 12 && \text{Rearrange} \end{aligned}

In other words, the son is aged 12, and since the father 22 years older, he must be 34. In 10 years, the son will be 22, and the father will be twice his age, 44. This problem is illustrated on the associated plot of the equations.

For other ways to solve this kind of equations, see below, System of linear equations.

Further reading

  1. Wikipedia, page 9710, revision 1357801777, Elementary algebra, section Solving algebraic equations > Linear equations with two variables. adopted
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