A17, A21 - Linear equations and mathematical models

Equality and inverse operations

A bill is £27, including £3 delivery and £4 for each identical box. How many boxes were bought? We need an unknown that makes the total correct, rather than a guess that happens to look reasonable.

An equation is like a balanced pair of quantities. Adding or subtracting the same amount on both sides preserves equality. So does multiplying or dividing both sides by the same non-zero number. Changing one side alone generally changes the solution.

A linear equation has the unknown only to the first power after simplification. A familiar form is $ax+b=c$. When $a\ne0$, inverse operations isolate $x$.

Change only the right side, from 20 to $-10$. The inverse operations stay the same.

A negative answer is allowed unless a context excludes it. A fractional coefficient works too:

Checking gives $-12/4-2=-5$. Division by zero is never an allowed solving step.

State the value of $x$ at the end. Stopping at $3x=15$ has not finished solving.

Brackets, fractions and both sides

Simplify first, then collect unknown terms on one side and constants on the other. Each “move” is really an equal operation on both sides.

For $3(x-2)=2x+7$, multiply both terms inside the bracket:

Both original sides then equal 33. The bracket protects the order of the calculation.

Numerical denominators can be cleared by multiplying every term by a common multiple. This makes collecting the unknown easier.

Change only the final constant on the right from $-1$ to $+1$. A fractional solution is perfectly valid.

With several denominators, use a common multiple. For

multiplying every term by 6 gives $2(x-1)+3x=24$. Expanding gives $5x-2=24$, then $x=\frac{26}{5}$. Substitution gives $\frac75+\frac{13}{5}=4$. This is multiplication of complete terms, not arbitrary removal of fraction bars.

Sometimes the unknown cancels. In $2(x+3)=2x+6$, simplification gives $6=6$, true for every real $x$. In $2(x+3)=2x+7$, it gives $6=7$, so there is no solution. A true identity does not force $x=0$.

An unknown in a denominator needs a restriction against zero; the examples here have fixed, non-zero numerical denominators.

Solutions as graph intersections

The two sides of an equation can be drawn as two heights. At an intersection, they have equal heights for the same input, so its horizontal coordinate solves the equation.

For $2x+1=6$, draw $y=2x+1$ and $y=6$ on the same axes. To construct the sloping line, calculate points such as $(0,1),(2,5),(4,9)$; two determine the line and a third checks the plotting. Use regular scales and a ruler. Draw the horizontal line through height 6, then read the intersection down to the $x$-axis.

[DIAGRAM: asset_name: A linear equation as two equal heights; asset_slug: 1ma1_m28_linear_solution; caption: The lines meet at (2.5, 6), so the equation has solution x = 2.5.; description: Coordinate axes show y = 2x + 1 in blue and y = 6 in clay, with a sage intersection at (2.5, 6) and a vertical reading guide to x = 2.5.]