C2.5 Equations

Learning objectives

C2.5.1Construct expressions, equations and formulas• Construct simple expressions, equations and formulas, including linear simultaneous equations.C2.5.2Solve linear equations• Solve linear equations in one unknown.C2.5.3Solve simultaneous linear equations• Solve simultaneous linear equations in two unknowns.C2.5.4Change the subject of simple formulas• Change the subject of a simple formula where the subject appears once and is not raised to a power or under a root.

Algebra translates quantities and relationships into symbols. Decide what each letter represents, express each quantity once, and use an equals sign only when two complete quantities have the same value.

Algebraic statement	Purpose	Example
expression	represents a quantity; no equals sign	2 more than $n$ : $n+2$
equation	states that two quantities are equal	$3n-5=22$
formula	defines one quantity using others	$C=56d+436$
simultaneous equations	two conditions on the same two unknowns	$3a+4c=38.5$ , $6a+5c=65$

Translate operations in their stated order. 'Three less than twice $x$ ' is $2x-3$ , while 'twice three less than $x$ ' is $2(x-3)$ . Brackets preserve which operation happens first.

Build a total by adding every contribution once. If Rovers score $x$ , United score $x+8$ , and City score $2x-3$ , a total of 117 gives $x+(x+8)+(2x-3)=117$ .

Use the same letters with the same meanings in both conditions. If a minibus costs $h$ dollars per hire-hour and $p$ dollars per passenger, the two trips give $3h+10p=61$ and $5h+8p=80$ .

Do not insert an equals sign into a requested expression, and do not solve unless asked. Keep units consistent: $3a+4c=38.5$ uses dollar prices, whereas cent prices would require 3850 on the right.

An equation is balanced because both sides have equal value. Any operation applied to one side must be applied to the other side; use inverse operations until the unknown is isolated.

Step	Action	Example for $12x-3=4x+21$
1	remove brackets or fractions if useful	none here
2	collect unknown terms on one side	$8x-3=21$
3	collect constants on the other side	$8x=24$
4	divide by the coefficient	$x=3$
5	substitute back to check	both sides equal 33

Either divide by an outside factor first or expand correctly. For $5(2x+4)=85$ , divide by 5 to get $2x+4=17$ , then $2x=13$ and $x=6.5$ .

Clear fractions by multiplying every term by a common denominator. For $3w/16-1=1/2$ , multiply all terms by 16: $3w-16=8$ , so $3w=24$ and $w=8$ .

When the unknown appears on both sides, move all unknown terms together before dividing. From $5x-6=x+3$ , subtract $x$ and add 6 to obtain $4x=9$ , so $x=2.25$ .

A term does not simply 'move and change sign'; the sign changes because the same inverse operation is performed on both sides. Never divide by the coefficient until every remaining term on that side belongs to the unknown term.

A simultaneous solution is one pair of values that makes both equations true at the same time. Elimination combines equivalent equations so that one unknown cancels.

Step	Action
1	align like terms and equals signs
2	multiply one or both equations so one pair of coefficients is equal or opposite
3	add or subtract the whole equations to eliminate that variable
4	solve the resulting one-variable equation
5	substitute back to find the second value and check both equations

Solve $5x-2y=44$ and $2x+3y=10$ . Multiply the first equation by 3 and the second by 2: $15x-6y=132$ and $4x+6y=20$ . Add to get $19x=152$ , so $x=8$ . Then $2(8)+3y=10$ , giving $y=-2$ .

Choose addition when coefficients are opposite and subtraction when they are equal. If neither is ready, multiply complete equations first; every term, including the right-hand side, must be multiplied.

In a context, define both unknowns and form two equations before eliminating. For adult price $x$ and child price $y$ , the conditions $2x+3y=15$ and $3x+5y=23.5$ use the same meanings and lead to one common price pair.

Finding values that satisfy only one equation is not enough. Substitute the final pair into both original equations. Do not add or subtract selected terms from an equation while leaving its other terms unchanged.

The subject of a formula is the single letter isolated on one side. To change the subject, preserve equality while undoing the operations around the required letter in reverse order.

Operation around the new subject	Apply to both sides
add $c$	subtract $c$
subtract $c$	add $c$
multiply by $m$	divide by $m$
divide by $m$	multiply by $m$

Make $x$ the subject of $y=mx+c$ . First subtract $c$ : $y-c=mx$ . Then divide by $m$ : $x=(y-c)/m$ , where $m
e0$. The numerator brackets show that the entire difference is divided.

Make $m$ the subject of $q=m/7+r$ . Subtract $r$ to get $q-r=m/7$ , then multiply by 7: $m=7(q-r)$ . Multiplying out to $7q-7r$ is equivalent.

For $5w-3y+7=0$ , isolate the term containing $w$ : $5w=3y-7$ . Divide by 5 to obtain $w=(3y-7)/5$ .

This Core objective uses formulas where the new subject appears once and is not under a power or root. Do not divide only one term of a numerator: $(y-c)/m$ means both $y$ and $-c$ are divided by $m$ .

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—C2.6 InequalitiesRead

C2.5 Equations

Learning objectives

Translate situations into expressions, equations and formulas

Solve linear equations by preserving balance

Solve simultaneous linear equations by eliminating one unknown

Change the subject using inverse operations

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