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SPI Mathematics · Formula sheet

Linear Equations formulas

6 formulas, 2 reference tables and 20 common traps for SPI Mathematics Linear Equations, grouped by subtopic.

Full notes with worked examples

Solving Linear Equations

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Solving an equation in one variable

One unknown on each side

ax+b=c  ⇒  x=c−baax+b=cx+d  ⇒  x=d−ba−c(a≠c)ax + b = c \;\Rightarrow\; x = \dfrac{c - b}{a} \qquad ax + b = cx + d \;\Rightarrow\; x = \dfrac{d - b}{a - c} \quad (a \ne c)

Solving a linear inequality

Flipping rule

a<b and k<0  ⇒  ka>kba < b \text{ and } k < 0 \;\Rightarrow\; ka > kb

Solving a pair of equations in two variables

Solution of a pair, and the solution check

x=c1b2−c2b1a1b2−a2b1,y=a1c2−a2c1a1b2−a2b1(x0,y0) solves ax+by=c when ax0+by0=cx = \dfrac{c_1b_2 - c_2b_1}{a_1b_2 - a_2b_1}, \quad y = \dfrac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1} \qquad (x_0, y_0) \text{ solves } ax + by = c \text{ when } ax_0 + by_0 = c

Common traps

Not changing the sign when a term moves

5x−2=3x+65x - 2 = 3x + 6 gives 2x=82x = 8, so x=4x = 4. Moving −2-2 across without changing its sign gives 2x=42x = 4 and a wrong x=2x = 2.

The minus sign in front of a fraction

−x−12-\frac{x - 1}{2} multiplied by 2 is −(x−1)=−x+1-(x - 1) = -x + 1, not −x−1-x - 1. Bracket the numerator before you multiply.

Multiplying only some of the terms

When clearing fractions, every term is multiplied, including a number standing alone. x3+2=5\frac{x}{3} + 2 = 5 times 3 is x+6=15x + 6 = 15, so x=9x = 9.

Forgetting to flip

−2x>6-2x > 6 gives x<−3x < -3. Dividing by −2-2 without turning the sign gives x>−3x > -3, which fails: at x=0x = 0, −2x=0-2x = 0 is not more than 6.

Flipping when the number is positive

Multiplying by 15 to clear the denominators 3 and 5 keeps the sign as it is, because 15 is positive. Only a negative multiplier flips it.

Checking in only one equation

A pair that fits one equation can miss the other. (5,2)(5, 2) fits x−y=3x - y = 3 but not x+2y=11x + 2y = 11, since 5+4=95 + 4 = 9.

Subtracting when the signs are opposite

If one equation has +6y+6y and the other −6y-6y, add the equations to remove yy. Subtracting them doubles the y-term instead.

Graphs and Consistency of a Pair

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Finding an unknown coefficient for a chosen case

Conditions on the coefficients

one: a1a2≠b1b2none: a1a2=b1b2≠c1c2infinitely many: a1a2=b1b2=c1c2\text{one: } \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \qquad \text{none: } \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \qquad \text{infinitely many: } \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

The graph of a linear equation

EquationGraphExample
ax+by=cax + by = c, with aa and bb not zeroA slanting straight line2x+y=42x + y = 4
x=ax = aA vertical line, parallel to the y-axisx=3x = 3
y=by = bA horizontal line, parallel to the x-axisy=5y = 5
x=0x = 0The y-axis itselfevery point (0,y)(0, y)
y=0y = 0The x-axis itselfevery point (x,0)(x, 0)
y=mxy = mxA line through the originy=2xy = 2x
x=ax = a stands up and y=by = b lies flat: each line is parallel to the axis of the letter that is missing from its equation.

The three cases of a pair of lines

RatiosLinesSolutionsPair is
a1a2≠b1b2\frac{a_1}{a_2} \ne \frac{b_1}{b_2}IntersectingExactly oneConsistent
a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}Coincident (the same line)Infinitely manyConsistent (dependent)
a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}ParallelNoneInconsistent
Two vertical lines such as x=1x = 1 and x=4x = 4 are parallel too, so a pair like that has no solution.
Compare the a and b ratios first. Only when they are equal does the c ratio decide.

Common traps

Getting the axis backwards

x=4x = 4 is parallel to the y-axis, not the x-axis. It is every point whose x is 4, and those points stand in an upright line.

Thinking a one-letter equation is not linear

y=3y = 3 is a linear equation in two variables, 0x+1y=30x + 1y = 3, and its graph is a straight line.

Comparing before clearing fractions

x2+y=3\frac{x}{2} + y = 3 and x+2y=6x + 2y = 6 look different, but doubling the first gives the second: they are the same line.

Calling coincident lines inconsistent

Coincident lines share every point, so the pair has infinitely many solutions and is consistent. Only parallel lines are inconsistent.

Keeping both roots of a square

A condition such as p2=25p^2 = 25 gives p=5p = 5 or p=−5p = -5. Test each in the third ratio: one of them usually gives parallel lines, not the same line.

Giving the excluded value as the answer

For a unique solution the answer is every value except the one that makes the first two ratios equal. That one value gives parallel or coincident lines.

Dropping a minus sign in a coefficient

In 3x−5y=73x - 5y = 7 the coefficient of yy is −5-5, not 5. Losing the sign gives the right size of answer with the wrong sign.

Word Problems on Linear Equations

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One unknown from a sentence

Fixed part plus a rate

y=a+bx(a=fixed part, b=rate per unit)x+(x+1)+(x+2)=3x+3y = a + bx \quad (a = \text{fixed part},\ b = \text{rate per unit}) \qquad x + (x + 1) + (x + 2) = 3x + 3

Chains of comparisons: ages and heights

Comparison sentences as equations

A=12B  ⇒  B=2AA=B+k (A is k older)s−m=t−s  ⇒  m+t=2s (s halfway)A = \tfrac{1}{2}B \;\Rightarrow\; B = 2A \qquad A = B + k \ (A \text{ is } k \text{ older}) \qquad s - m = t - s \;\Rightarrow\; m + t = 2s \ (s \text{ halfway})

Common traps

Stopping at x

If x is the smallest piece and the question asks for the longest, work out the longest piece's expression. The value of x itself is often printed as a wrong option.

Reversing 'more than'

'A has 5 more than B' is A=B+5A = B + 5, not B=A+5B = A + 5. Decide who has more before you write the equation.

Mixing metres and centimetres

2.5 m is 250 cm. Adding 30 cm to a height still written as 2.5 gives nonsense, so change units first.

Reading 'half that of' backwards

'P's age is half that of Q' means P=Q2P = \frac{Q}{2}, so Q is the older. Writing Q=P2Q = \frac{P}{2} swaps the two people.

Using one equation for two facts

The number of coins and their value are two different facts. One equation adds the coins; the other adds what each coin is worth.

Adding to the top only

'1 is added to both the numerator and the denominator' changes xy\frac{x}{y} to x+1y+1\frac{x + 1}{y + 1}. Adding to the numerator alone answers a different question.

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