SAT

Linear equations in one variable

A linear equation in one variable can be rearranged to $ax + b = c$, and solving it means undoing what has been done to $x$ in the reverse order: clear brackets and fractions first, collect the $x$ terms on one side, then divide.

Solve $3(2x - 5) + 4 = 5x + 1$.

Checking in the original: the left side is $3(24 - 5) + 4 = 61$ and the right side is $60 + 1 = 61$.

When an equation has fractions, multiplying every term by a common denominator removes them. For $\frac{x}{4} + \frac{x}{6} = 5$ the common denominator is 12, which gives $3x + 2x = 60$ and so $x = 12$.

How many solutions

Take an equation of the form $ax + b = cx + d$ and collect terms to get $(a - c)x = d - b$. There are three cases.

This is how questions with an unknown constant are solved. If $4(x + 3) = kx + 12$ has infinitely many solutions, the left side is $4x + 12$, so the coefficients of $x$ must match and $k = 4$. If $6x + 5 = kx + 9$ has no solution, the $x$ coefficients must match while the constants differ, so $k = 6$.

Rearranging a formula

Solving for one letter in terms of the others uses the same steps, treating the other letters as numbers. From $P = 2\ell + 2w$, subtract $2\ell$ and divide by 2 to get $w = \frac{P - 2\ell}{2}$.

Linear equations in two variables

An equation such as $y = 3x - 2$ has infinitely many solutions, each a pair $(x, y)$, and they lie on a straight line. There are three common forms of the same line.

The slope between two points is

It is the change in $y$ for each increase of 1 in $x$. A horizontal line $y = c$ has slope 0, and a vertical line $x = c$ has an undefined slope.

Finding a line's equation

For the points $(2, 7)$ and $(6, 19)$ the slope is $\frac{12}{4} = 3$. Point-slope form with $(2, 7)$ gives $y - 7 = 3(x - 2)$, so $y = 3x + 1$. The point $(6, 19)$ checks: $3 \times 6 + 1 = 19$.

Standard form

For $Ax + By = C$ the slope is $-\frac{A}{B}$, the $y$-intercept is $\frac{C}{B}$ and the $x$-intercept is $\frac{C}{A}$. For $3x + 4y = 24$ the slope is $-\frac{3}{4}$, the $x$-intercept is 8 and the $y$-intercept is 6.

Parallel and perpendicular lines

Parallel lines have equal slopes. Perpendicular lines have slopes whose product is $-1$, so one slope is the negative reciprocal of the other.

The line through $(4, 1)$ perpendicular to $y = 2x + 5$ has slope $-\frac{1}{2}$, because $2 \times -\frac{1}{2} = -1$. Then $y - 1 = -\frac{1}{2}(x - 4)$, which simplifies to $y = -\frac{1}{2}x + 3$. At $x = 4$ this gives $-2 + 3 = 1$, as required.

Linear functions and models

A linear function is written $f(x) = mx + b$. The notation $f(4)$ means the output when $x = 4$, and the graph of $y = f(x)$ is a line, so a point $(a, b)$ on the graph means $f(a) = b$.

If $f(3) = 11$ and $f(7) = 23$, the slope is $\frac{23 - 11}{7 - 3} = 3$ and $f(3) = 3 \times 3 + b = 11$ gives $b = 2$, so $f(x) = 3x + 2$. To find the input that gives an output of 50, solve $3x + 2 = 50$, which gives $x = 16$.

Reading a model in context

In a linear model the slope is the rate of change, stated with units of output per unit of input, and the intercept is the value when the input is 0. A candle 20 cm tall that burns at 1.5 cm per hour is modelled by $h = 20 - 1.5t$, where $t$ is the time in hours.

• The slope $-1.5$ means the height falls by 1.5 cm each hour.
• The $h$-intercept 20 is the starting height.
• The $t$-intercept comes from $0 = 20 - 1.5t$, so $t = \frac{40}{3} \approx 13.3$ hours, when the candle has burned down.

Building the equation from a description works the same way. A fixed amount goes in as the intercept and an amount that repeats for each unit goes in as the slope. A gym with a 40 dollar joining fee and 25 dollars per month gives $C = 25m + 40$ after $m$ months. If a question gives a total and asks for the number of units, substitute the total and solve.

Direct variation is the special case $y = kx$, a line through the origin, so doubling $x$ doubles $y$.