Equations and Inequalities

A linear equation is a balance: do the same thing to both sides

An equation is a statement that two expressions are equal, and it is true only for particular values of the letter. $3x+4=10$ is true when $x=2$ and false for every other value, so $x=2$ is the solution. Solving means finding that value, and the whole method rests on one rule: whatever you do to one side of an equation you must do to the other side, so the balance is never disturbed.

Undo the operations in reverse order. In $3x+4=10$ the letter was multiplied by 3 and then had 4 added, so you subtract 4 first and divide by 3 second: $3x+4=10$ gives $3x=6$, which gives $x=2$. In $x+7=12$ there is only one operation to undo, so $x=5$. Where the letter already stands alone with a coefficient, as in $4x=20$, divide both sides: $x=5$.

When the letter appears on both sides, collect the letter terms on the side where the coefficient ends up positive, because that keeps the arithmetic free of minus signs. $5-2x=3(x+7)$ expands to $5-2x=3x+21$. Adding $2x$ to both sides gives $5=5x+21$, so $5x=-16$ and $x=-3.2$. Negative and fractional answers are normal; a tidy-looking answer is not evidence of a correct one.

Finish by substituting your answer into the equation exactly as the question printed it, never into the line you finished on. For $x=-3.2$: $5-2(-3.2)=11.4$, and $3(-3.2+7)=3\times 3.8=11.4$, so the two sides agree.

Brackets and fractions: the same rules, tidier working

Brackets are not a new kind of equation. Expand them and you are back to a linear equation you already know how to solve. $4(x-3)=2x+6$ becomes $4x-12=2x+6$, then $2x=18$ and $x=9$. Check it: $4(9-3)=24$ and $2(9)+6=24$.

An equation with numerical fractions is cleared by multiplying every term by a common multiple of the denominators. For $\frac{x}{3}+\frac{x}{4}=7$, the LCM of 3 and 4 is 12, so multiplying every term by 12 gives $4x+3x=84$, then $7x=84$ and $x=12$. Multiplying only the first term is the mistake that costs the mark - the 7 has to be multiplied by 12 as well.

Extended only: fractional equations can also carry algebraic denominators, such as $\frac{2}{x+2}+\frac{3}{2x-1}=1$. Multiply every term by the product of the two denominators, $(x+2)(2x-1)$, which gives $2(2x-1)+3(x+2)=(x+2)(2x-1)$. Working both sides out gives $7x+4=2x^2+3x-2$, so $2x^2-4x-6=0$ and $x^2-2x-3=0$. That factorises as $(x-3)(x+1)=0$, so $x=3$ or $x=-1$.

Extended only: before you multiply, note the values of $x$ that would make a denominator zero - here $x=-2$ and $x=0.5$. Those values can never be solutions, so checking them is part of the method rather than an optional extra.

• $\frac{2x-1}{5}=3$: multiply by 5, so $2x-1=15$ and $x=8$
• $\frac{x}{3}+\frac{x}{4}=7$: multiply every term by 12, so $4x+3x=84$ and $x=12$
• Extended only: $\frac{x}{2x+1}=4$: multiply by $2x+1$, so $x=8x+4$ and $x=-\frac{4}{7}$

Changing the subject of a formula

The subject of a formula is the letter that stands on its own, usually on the left: in $v=u+at$ the subject is $v$. Changing the subject means rearranging so that a different letter stands alone, and the balance rule is exactly the one you use for equations. The only difference is that the other letters travel with the numbers instead of being replaced by them.

Work backwards through the formula, undoing the operations that were applied to the letter you want. To make $x$ the subject of $y=5x-3$, the letter was multiplied by 5 and then had 3 subtracted, so add 3 and then divide by 5: $y+3=5x$, so $x=\frac{y+3}{5}$. Where the formula contains a fraction, multiply by the denominator: from $A=\frac{1}{2}bh$, multiplying by 2 gives $2A=bh$ and so $h=\frac{2A}{b}$.

Extended only: when the subject appears twice, collect every term containing it on one side, factorise it out, then divide. To make $x$ the subject of $y=\frac{2x+1}{x-3}$: multiply by the denominator, $y(x-3)=2x+1$; expand, $yx-3y=2x+1$; collect the $x$ terms on one side, $yx-2x=3y+1$; factorise, $x(y-2)=3y+1$; divide, $x=\frac{3y+1}{y-2}$.

Extended only: where there is a power or a root of the subject, undo it with the opposite operation. To make $b$ the subject of $c=\sqrt{a^2+b^2}$: square both sides, $c^2=a^2+b^2$; subtract, $b^2=c^2-a^2$; take the square root, $b=\sqrt{c^2-a^2}$.

• Core: the subject appears once and there is no power or root of the subject
• Extended only: the subject appears twice - collect the terms, factorise, then divide
• Extended only: there is a power or root of the subject - square both sides, or take the root of both sides

Simultaneous equations by elimination

Two equations in two unknowns are called simultaneous equations, and one pair of values has to satisfy both of them at the same time. Write the equations one under the other with the same letters in the same columns. That single habit prevents most of the errors in this topic, because a term lined up in the wrong column is silently added to the wrong letter.

Elimination removes one letter by making its coefficients the same size and then adding or subtracting the equations. Solve $3x+2y=17$ with $5x-2y=7$: the $y$ terms are $+2y$ and $-2y$, so adding the equations cancels them. $8x=24$, so $x=3$. Substitute that into the first equation: $9+2y=17$, so $y=4$. The solution is $x=3$, $y=4$.

If the signs of the chosen letter are different, add the equations; if the signs are the same, subtract one equation from the other. When the coefficients do not match, multiply one or both equations first. For $3c+2t=11.40$ and $2c+3t=11.10$, multiply the first by 3 and the second by 2 to get $9c+6t=34.20$ and $4c+6t=22.20$; subtracting gives $5c=12.00$, so $c=2.40$, and substituting back gives $t=2.10$.

Core also expects you to construct the equations from a description. Read the description, name the two unknowns in words, and write one equation for each total it gives you. Two totals give you the two equations you need.

Simultaneous equations by substitution, and one linear with one non-linear

Substitution rearranges one equation so that a letter stands alone, then puts that expression into the other equation. Solve $y=3x-1$ with $2x+3y=19$: replacing $y$ gives $2x+3(3x-1)=19$, so $11x-3=19$, $x=2$ and $y=5$.

Choose the quicker method and do not worry about which one the examiner wanted. Substitution is quicker when one letter already has coefficient 1, as in $y=3x-1$ or $x+2y=7$. Elimination is quicker when the coefficients of one letter are equal or one is a multiple of the other. Either method scores full marks.

Extended only: one linear equation with one non-linear equation is solved by substitution. Substitute the linear expression into the non-linear one, which turns it into a quadratic, solve that, then find the matching value of the other letter for each root. Solve $y=x+1$ with $x^2+y^2=25$: substituting gives $x^2+(x+1{)}^2=25$, so $2x^2+2x-24=0$ and $x^2+x-12=0$. Factorising gives $(x+4)(x-3)=0$, so $x=-4$ or $x=3$, and the matching values are $y=-3$ and $y=4$. The two solutions are the pairs $(-4,-3)$ and $(3,4)$.

Extended only: check each pair in both equations before you write them down. A pair that satisfies the quadratic but not the straight line is the classic result of a sign slip in the substitution.

• $y=3x-1$ with $2x+3y=19$ gives $x=2$, $y=5$
• Extended only: $y=x+1$ with $x^2+y^2=25$ gives the pairs $(-4,-3)$ and $(3,4)$
• A straight line and a curve normally meet in two points, so expect two pairs of answers and give both

Quadratic equations by factorising (Extended only)

A quadratic equation contains $x^2$, and its standard form is $a{x}^2+bx+c=0$ with everything on one side. Rearranging is not optional: $x^2=5x-6$ has to become $x^2-5x+6=0$ before anything else can happen to it.

Factorising reverses the expansion of two brackets. You need two numbers that multiply to give $c$ and add to give $b$. For $x^2-5x+6$ those numbers are $-2$ and $-3$, so $x^2-5x+6=(x-2)(x-3)$. The method then rests on the zero rule: if a product of two numbers is zero, at least one of the factors must be zero. So $x-2=0$ or $x-3=0$, giving $x=2$ or $x=3$.

When the coefficient of $x^2$ is not 1, expect one bracket to begin with $ax$. $2x^2+5x-3=(2x-1)(x+3)$, which expands to $2x^2+6x-x-3=2x^2+5x-3$. So $2x-1=0$ or $x+3=0$, giving $x=\frac{1}{2}$ or $x=-3$. A difference of two squares factorises in one line: $x^2-49=(x-7)(x+7)$.

Never divide a quadratic equation through by $x$ to make it look simpler. $x^2=5x$ has two solutions, and dividing by $x$ throws one of them away. Rearrange to $x^2-5x=0$, factorise to $x(x-5)=0$, and both roots $x=0$ and $x=5$ appear.

The quadratic formula and completing the square (Extended only)

The quadratic formula solves any quadratic equation, whether or not it factorises. For $a{x}^2+bx+c=0$, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. It is printed in the List of formulas, so you do not have to memorise it, but you must write down $a$, $b$ and $c$ with their signs before you substitute anything.

For $2x^2-7x+3=0$, $a=2$, $b=-7$ and $c=3$, so $x=\frac{7\pm \sqrt{49-24}}{4}=\frac{7\pm 5}{4}$, giving $x=3$ or $x=0.5$. Taking the whole of $-b$ means taking $+7$; writing $b=7$ and then $x=\frac{-7\pm 5}{4}$ is the commonest way to lose both roots at once.

When $b^2-4ac$ is not a perfect square the answer is a surd, which can be left exact or rounded as the question asks. For $x^2-4x+1=0$: $x=\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}$, because $\sqrt{12}=2\sqrt{3}$. If $b^2-4ac$ is negative there is no real square root, so the equation has no real solutions.

Completing the square writes the same expression as a squared bracket plus a constant: $x^2+bx+c=(x+\frac{b}{2}{)}^2-(\frac{b}{2}{)}^2+c$. So $x^2+6x+1=(x+3{)}^2-9+1=(x+3{)}^2-8$. Setting that to zero gives $(x+3{)}^2=8$, so $x+3=\pm \sqrt{8}$ and $x=-3\pm 2\sqrt{2}$, which is $-0.17$ or $-5.83$ to two decimal places. The completed square form also reads off the turning point of the curve $y=x^2+6x+1$: it is $(-3,-8)$.

• $2x^2-7x+3=0$: $a=2$, $b=-7$, $c=3$, so $x=\frac{7\pm 5}{4}$, giving $x=3$ or $x=0.5$
• $x^2-4x+1=0$: $x=2\pm \sqrt{3}$, an exact surd answer
• $x^2+6x+1=0$: $(x+3{)}^2=8$, so $x=-3\pm 2\sqrt{2}$

Inequalities: number lines, solving and regions

An inequality compares two expressions using $<$, $>$, $\le$ or $\ge$. Solving one follows the balance rule, with one extra rule: multiplying or dividing by a negative number reverses the inequality sign. From $-2x<8$, dividing both sides by $-2$ gives $x>-4$. Check it with a value that should be allowed: $x=0$ gives $-2\times 0=0$, and $0<8$, so $0$ is allowed - and $0$ is greater than $-4$, exactly as the answer says.

On a number line an open circle marks a value that is not included, and a closed circle marks one that is. So $-3\le x<1$ is drawn with a closed circle at $-3$, an open circle at 1, and a thick line joining them. Reading a number line works the same way backwards: a closed circle at 2 with the line running to the right is $x\ge 2$.

Extended only: solving linear inequalities, including the two-ended kind. $-3\le 3x-2<7$ is solved by doing the same thing to all three parts: add 2 to each part, $-1\le 3x<9$, then divide each part by 3, $-\frac{1}{3}\le x<3$. A one-ended example is $3x<2x+4$, where subtracting $2x$ from both sides gives $x<4$.

Extended only: an inequality in two variables such as $y\le 2x+1$ is drawn as a region. Sketch the boundary line $y=2x+1$, using a solid line for $\le$ or $\ge$ and a broken line for $<$ or $>$, then shade the region you do not want - that is the convention unless the question says otherwise. Test one point to decide which side is which: $(0,0)$ gives $0\le 1$, which is true, so the wanted region is the one containing the origin. Listing the inequalities that define a shaded region is the same work in reverse: find each boundary line, then read off its inequality and whether the line is solid or broken.

Extended only: a quadratic inequality is solved from the roots. $x^2-2x-3=0$ has roots $x=-1$ and $x=3$, and the curve opens upwards, so the curve is below the x-axis between the roots. That gives $x^2-2x-3<0$ for $-1<x<3$, and $x^2-2x-3>0$ for $x<-1$ or $x>3$.

• $3x<2x+4$: subtract $2x$, so $x<4$
• $-3\le 3x-2<7$: $-\frac{1}{3}\le x<3$
• $x^2-2x-3<0$: $-1<x<3$, while $x^2-2x-3>0$: $x<-1$ or $x>3$
• $y\le 2x+1$: solid line, shade the unwanted region above it; $y>2x+1$: broken line, shade below