6.02b, 6.02e - Substituting into formulae | OCR GCSE Mathematics

Can the same formula work with positive and negative inputs? Substitute values into expressions and formulae, keeping signs, powers, roots and fractions in the right order, then use supplied formulae for motion with constant acceleration.

Substituting positive values

A hire charge is $C=5+3h$ , where $C$ is the cost in pounds and $h$ is the number of hours. What changes when the hire lasts four hours rather than two? The rule stays fixed; you replace the input.

Substitution

To substitute, replace every occurrence of a variable with its given value, then calculate. In a formula written as $C=5+3h$ , $C$ is the subject: it is already alone on one side.

Write the substituted calculation before evaluating it. Remember that $3h$ means $3\times h$ , and that multiplication is done before addition. A formula is a rule, not an instruction to read operations from left to right regardless of priority.

Worked example 1: a positive input

Question. Find $C$ when $C=5+3h$ and $h=2$ .

Plan. Replace $h$ with two and multiply before adding.

\begin{aligned} C&=5+3\times2 &&\text{substitute }h=2\\ &=5+6 &&\text{multiply first}\\ &=11 &&\text{add the fixed amount} \end{aligned}

Answer. $C=11$ , so the cost is £11.

Check. Two hourly charges of £3 total £6; adding the £5 fixed charge gives £11.

Change only the input to four hours. The fixed charge remains five pounds.

Worked example 2: a larger input

Question. Find $C$ when $C=5+3h$ and $h=4$ .

Plan. Use the same rule with the new value of $h$ .

\begin{aligned} C&=5+3\times4 &&\text{replace the variable}\\ &=5+12 &&\text{calculate the product}\\ &=17 &&\text{add the fixed amount} \end{aligned}

Answer. $C=17$ , so the cost is £17.

Check. Two extra hours should add $2\times£3=£6$ . The answer increased from £11 to £17, as expected.

In an expression such as $ab$ , substitute both values with a multiplication sign: if $a=2$ , $b=7$ , then $ab=2\times7=14$ , not $27$ .

Watch out

$5+3\times4$ is $17$ , not $32$ . Adding five and three first would change the formula into $(5+3)h$ .

Use a new fixed charge and hourly rate.

Substituting negative values

A negative input needs a visible boundary. Substituting $x=-3$ into $2x$ gives $2\times(-3)$ . Brackets show that the entire negative number replaces $x$ .

Keep the formula's operations separate from the input's sign. In $7-2x$ , the subtraction is still present when $x$ is negative. For $x=-3$ , the product $2x$ is $-6$ , so you subtract negative six.

You already know

Subtracting a negative number means adding its opposite. A negative input is not an instruction to change every sign in the formula.

Work through the product first, keeping the outer subtraction until the next line.

Worked example 3: one negative input

Question. Find $R$ when $R=7-2x+y$ , $x=-3$ and $y=4$ .

Plan. Replace both letters and evaluate the signed product before the sum.

\begin{aligned} R&=7-2(-3)+4 &&\text{substitute with brackets}\\ &=7-(-6)+4 &&\text{multiply first}\\ &=7+6+4=17 &&\text{subtract the negative} \end{aligned}

Answer. $R=17$ .

Check. The $x$ -contribution adds six to seven, then the $y$ -contribution adds four: $13+4=17$ .

Change only the sign of $y$ . The $x$ -contribution stays the same.

Worked example 4: two negative inputs

Question. Find $R$ when $R=7-2x+y$ , $x=-3$ and $y=-4$ .

Plan. Substitute each signed value independently.

\begin{aligned} R&=7-2(-3)+(-4) &&\text{replace both variables}\\ &=7+6-4 &&\text{apply the sign rules}\\ &=9 &&\text{combine the changes} \end{aligned}

Answer. $R=9$ .

Check. Changing $y$ from $4$ to $-4$ reduces the answer by eight. The previous result was seventeen, and $17-8=9$ .

If a letter occurs twice, replace it twice. For example, $x+x^2$ at $x=-2$ becomes $(-2)+(-2)^2$ , not $(-2)+2^2$ by guesswork. Here both versions happen to agree because a square removes the sign; that coincidence will fail for an odd power.

Watch out

The line $7-2(-3)=7-6$ loses the sign of the product. Write $2(-3)=-6$ first, then $7-(-6)=13$ .

Keep both negative inputs visible on your first line.

Powers, roots and fractions in formulae

The square of a negative number is positive: $(-4)^2=16$ . But $-4^2=-16$ , because the power is evaluated before the minus outside it. Substitution brackets tell you which calculation the formula needs.

For a fraction, calculate the whole numerator and the whole denominator before dividing. A denominator of zero makes the expression undefined. For a real square root, the value under the root must be non-negative. The symbol $\sqrt{a}$ means the non-negative square root of $a$ , so $\sqrt{16}=4$ , not $\pm4$ .

Square-root and division restrictions

In real-number calculations, $\sqrt{A}$ requires $A\ge0$ , and $\frac{A}{B}$ requires $B\ne0$ . Check these conditions after substitution, before evaluating.

Notice how brackets around a negative input preserve its whole value when it is squared.

Worked example 5: a power, a fraction and a root

Question. Find $F$ when $F=\sqrt{\frac{x^2+11}{y-1}}$ , $x=5$ and $y=5$ .

Plan. Evaluate the squared input, then the fraction, then the square root.

\begin{aligned} F&=\sqrt{\frac{5^2+11}{5-1}} &&\text{substitute both values}\\ &=\sqrt{\frac{25+11}{4}} &&\text{power and denominator}\\ &=\sqrt9=3 &&\text{divide, then take the root} \end{aligned}

Answer. $F=3$ .

Check. The denominator is four, not zero, and the root input is nine, which is non-negative. Also $3^2\times4=36=25+11$ , reversing the operations.

Change only $x$ from five to negative five. Its square will stay the same, so predict the answer before calculating.

Worked example 6: a negative input under a square

Question. Find $F$ when $F=\sqrt{\frac{x^2+11}{y-1}}$ , $x=-5$ and $y=5$ .

Plan. Put the negative input in brackets before squaring it.

\begin{aligned} F&=\sqrt{\frac{(-5)^2+11}{5-1}} &&\text{keep the signed base}\\ &=\sqrt{\frac{25+11}{4}} &&\text{square the entire input}\\ &=\sqrt9=3 &&\text{fraction before root} \end{aligned}

Answer. $F=3$ .

Check. $(-5)(-5)=25$ , exactly as $5\times5=25$ . The denominator and every other operation are unchanged, explaining the identical result.

This answers the opening question: the same formula can accept either sign here, because the square gives the same value. It cannot accept every input: $y=1$ would make the denominator zero. With $x=5$ , $y=0$ , the fraction would be $-36$ , which has no real square root.

Watch out

Do not write $(-5)^2=-25$ or cancel the root with only one term of a sum. The square is $25$ , and the whole value beneath the root must be calculated before taking its root.

Try the same order with fresh values.

Using supplied motion formulae

Formulae can connect a moving object's initial velocity, final velocity and displacement. The substitution method is unchanged; first read what each letter means in the question.

Constant-acceleration motion

For motion with constant acceleration,

v=u+at\qquad s=ut+\frac12at^2\qquad v^2=u^2+2as

Here $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration, $t$ is elapsed time and $s$ is displacement from the starting position. With metres and seconds, velocities are in $\mathrm{m/s}$ and acceleration is in $\mathrm{m/s^2}$ .

These three formulae are not on the 2027 formula sheets. Learn to use them; the question supplies the meanings of their letters, which you are not required to recall.

Use $v=u+at$ when time is given and final velocity is wanted. Use $s=ut+\frac12at^2$ to find displacement from time. Use $v^2=u^2+2as$ when time is absent, or as a check linking the other quantities. To find a forward velocity from a positive value of $v^2$ , take its positive square root. The squared formula alone does not decide the sign: $v^2=25$ allows $v=5$ or $v=-5$ , so the direction information matters.

Acceleration and velocity can be signed. A negative acceleration reduces a positive velocity, provided the object has not yet reversed direction. In the following examples it keeps moving forwards, so displacement equals distance travelled.

Worked example 7: speeding up

Question. An object moves forwards with initial velocity $u=8\,\mathrm{m/s}$ and constant acceleration $a=1\,\mathrm{m/s^2}$ for $t=3\,\mathrm{s}$ . Find its final velocity $v$ and displacement $s$ . Check them using $v^2=u^2+2as$ .

Plan. Substitute into the first two formulae, then compare both sides of the third.

\begin{aligned} v&=8+1\times3=11 &&\text{initial value plus change}\\ s&=8\times3+\tfrac12\times1\times3^2 &&\text{square time first}\\ &=24+4.5=28.5 &&\text{combine the contributions} \end{aligned}

Answer. $v=11\,\mathrm{m/s}$ and $s=28.5\,\mathrm{m}$ .

Check. $v^2=11^2=121$ , while $u^2+2as=8^2+2\times1\times28.5=64+57=121$ . Both sides agree.

Change only the sign of the acceleration. The elapsed time is still squared before multiplication; the minus sign belongs to $a$ , not to $t$ .

Worked example 8: slowing down

Question. The same object has $u=8\,\mathrm{m/s}$ , $a=-1\,\mathrm{m/s^2}$ and $t=3\,\mathrm{s}$ . It continues forwards throughout. Find $v$ and $s$ , then check using $v^2=u^2+2as$ .

Plan. Substitute the negative acceleration in brackets in each formula.

\begin{aligned} v&=8+(-1)\times3=5 &&\text{velocity decreases}\\ s&=8\times3+\tfrac12\times(-1)\times3^2 &&\text{keep the signed factor}\\ &=24-4.5=19.5 &&\text{subtract its contribution} \end{aligned}

Answer. $v=5\,\mathrm{m/s}$ and $s=19.5\,\mathrm{m}$ .

Check. $v^2=25$ , and $u^2+2as=64+2\times(-1)\times19.5=64-39=25$ . A forward velocity must be positive, so the squared formula also gives $v=\sqrt{25}=5\,\mathrm{m/s}$ .

Both journeys last three seconds, yet the second covers less distance because it slows down. A negative acceleration has not made the displacement negative here: the object still travels forwards.

Watch out

In $\frac12at^2$ , square $t$ only. With $a=-1$ , $t=3$ , this term is $\frac12(-1)\times9=-4.5$ , not $\frac12(-3)^2=4.5$ .

Read every supplied meaning before using the same three formulae.

Quick Recap

Replace every occurrence of a variable with its given value.
Bracket negative inputs and keep the formula's operation signs.
Evaluate powers, complete numerators and denominators, then divisions and roots in the required order.
Check non-zero denominators and non-negative real square-root inputs.
Read supplied motion formulae and letter meanings, and use them only for constant acceleration.

A formula can work for either sign of an input, but its operations and restrictions decide which inputs are allowed and whether changing a sign changes the result.

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6.02b, 6.02e - Substituting into formulae

Substituting positive values

Substituting negative values

Powers, roots and fractions in formulae

Using supplied motion formulae

Quick Recap

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