fbpx

Algebra Foundation Revision Card Answers

FA1 – Algebra Basics & Collecting Like Terms

QUESTION: Simplify the following expression.

3p+3p+3pt-6pt+7p

ANSWER: Firstly, collect the three like terms 3p, 3p, and 7p to get

3p+3p+7p=13p

Then, collect the like terms 3pt and -6pt to get

3pt-6pt=-3pt

Therefore, the fully simplified expression is

13p-3pt

FA2 – Simplifying & Solving Equations

QUESTION: Solve the equation 5x+18=2-3x.

ANSWER: We want the x terms on one side. So, adding 3x to both sides, we get

8x+18=2

Then, subtracting 18 from both sides, we get

8x=-16

Finally, dividing both sides by 8, we get

x=\dfrac{-16}{8}=-2

Note: it’s possible to take a different route but get to the same answer. As long as your steps are mathematically correct, and your answer is -2, then you get full marks.

FA3 – Forming & Solving Equations

QUESTION: At the cinema, one large drink costs £4 and one adult ticket costs £x. Laline buys 2 adult tickets and one large drink.

         a) Write an expression, in terms of x, for the total cost of Laline’s cinema trip.

         b) The total cost of Laline’s trip was £26. Use your answer to part a) to form an equation, and

              use that equation to determine the value of x.

ANSWER: a) Laline bought two tickets, so that’s two lots of £x, and one large drink, which is one lot of £4. Adding these together, we get the expression

2x+4

b) The total cost of her trip was £26, and we know the expression above also represents the total cost of her trip, so we can set them equal to each other. Doing so, we get

2x+4=26

We can now solve this equation. Subtracting 4 from both sides, we get

2x=26-4=22

Then, dividing both sides by 2 gives us the answer

x=22\div2=\pounds 11.

FA4 – Rules of Indices

QUESTION: Evaluate the following expression. (non-calculator)

\dfrac{\left(5^4\right)^4}{5^{24}\times 5^{-6}}

ANSWER: First up, the numerator. The term has a power on top of another power, so they should be multiplied to get

\left(5^4\right)^4=5^{4\times4}=5^{16}

Then, the denominator. The terms are being multiplied, so we add the powers. Note: we are going to add a positive number to a negative number in exactly the same we always add a positive number to a negative number. Doing so, we get

5^{24}\times 5^{-6}=5^{24+\left(-6\right)}=5^{18}

Next, we can treat the whole fraction as a division (meaning we will subtract the powers). So, after having simplified the top and bottom we get

\dfrac{5^{16}}{5^{18}}=5^{16-18}=5^{-2}

This is a negative power, so it becomes

5^{-2}=\dfrac{1}{5^2}=\dfrac{1}{25}

Thus, we have evaluated the expression to be \frac{1}{25}.

FA5 – Rearranging Formulae

QUESTION: Rearrange the following formula to make t the subject.

v=\dfrac{1}{2}at^2

ANSWER: Firstly, we will multiply both sides by 2 to get

2v=at^2

Then, dividing both sides by a gives us

\dfrac{2v}{a}=t^2

We are almost there but not quite – we need to get rid of the power of 2. To do that, we must square root both sides to get

\sqrt{\dfrac{2v}{a}}=t

Flipped around (this step is not necessary, but is good practice), this looks like

t=\sqrt{\dfrac{2v}{a}}

FA6 – Expanding Single Brackets

QUESTION: Expand pqr(5pr+5r^5-25pqr).

ANSWER: We must multiply the term on the outside, pqr, by all the terms on the inside. To do this, we must use the law of indices that says: when you multiply terms, their powers are added. So, multiplying the terms out, we get

\begin{aligned}pqr\left(5pr+5r^5-25pqr\right)&=\left(pqr\times 5pr\right)+\left(pqr\times 5r^5\right)+\left(pqr\times \left(-25pqr\right)\right) \\ &=5p^{2}qr^2+5pqr^6-25p^{2}q^{2}r^{2}\end{aligned}

FA7 – Factorising Single Brackets

QUESTION: Factorise 8fg+fgh-4f^{3}gh.

ANSWER: We are looking for factors shared by all 3 terms. In this case, there is no number they are all divisible by, but they do share factors of f and g. So, taking these factors out in turn, we get

\begin{aligned}8fg+fgh-4f^{3}gh&=f\left(8g+gh-4f^{2}gh\right) \\ &=fg\left(8+h-4f^{2}h\right)\end{aligned}

The terms in the bracket have no common factors, so we are done.

FA8 – Expanding Double Brackets

QUESTION: Expand and simplify (y-6)(y-2).

ANSWER: We will apply FOIL, drawing red lines as we go, and then collect like terms once the expansion is done. So, we get

FA9 – Factorising Quadratics

QUESTION: Factorise k^2-7k+10.

ANSWER: We want two numbers which multiply to make 10 and add to make -7. In order for them to multiply to make a positive number but add to make a negative one, we need them to both be negative. So, considering factors of 10,

10=(-1)\times(-10)=(-2)\times(-5)

We see that (-2)\times(-5)=10 and -2+(-5)=-7, the two numbers we want are -2 and -5. Therefore, the quadratic factorises to

k^2-7k+10=(k-2)(k-5)

FA10 – Solving Quadratics by Factorisation

QUESTION: Solve z^2-2z-24=0.

ANSWER: We want two numbers which multiply to make -24 and add to make -2. So, considering factors of -24,

-24=1\times(-24)=2\times(-12)=3\times(-8)=4\times(-6)

We see that 4\times(-6)=-24 and 4+(-6)=-2, so the two numbers we want are 4 and -6. Therefore, the equation becomes

(z-6)(z+4)=0

Therefore, the two solutions are

z=6\,\,\,\,\text{and}\,\,\,\,z=-4

FA11 – Linear Sequences & the nth Term

QUESTION: Find the nth term of the sequence below.

1,\,\,\,5,\,\,\,9,\,\,\,13,\,\,\,17

ANSWER: The formula must take the form an+b. To find a, find the common difference between the terms and confirm that they are the same.

So, the formula must be 4n+b. So, we now write out the sequence given by 4n (i.e., the 4 times table):

4,\,\,\,8,\,\,\,12,\,\,\,16,\,\,\,20

Each of these terms is 3 bigger than their respective terms in the original sequence, so we must subtract 3 from them all. Therefore, the nth term is

4n-3

FA12 – Inequalities on a Number Line

QUESTION: On a number line, display the following inequality.

1\leq x\leq 8

ANSWER: There are boundaries at 1 and 8, so we will draw two circles there on the number. They are both inclusive inequalities, so we will draw closed (filled-in) circles. Then, since the only numbers included in the inequality are those in between the two boundaries, we connect the two circles with a line. This looks like

FA13 – Solving Inequalities

QUESTION: Solve the following inequality and display your answer on a number line.

2-4z\leq z-18

ANSWER: We will rearrange to make z the subject. So, adding 4z to both sides, we get

2\leq 5z-18

Then, adding 18 to both sides, we get

20\leq 5z

Finally, dividing both sides by 5 gives us the solution

4\leq z

So now we need to draw a closed circle at 4 on the number line (the inequality is inclusive), and draw an arrow going to the right, away from the circle. This looks like

FA14 – Simultaneous Equations

QUESTION: Solve the following simultaneous equations.

\begin{aligned}2a-4b&=-1 \\ 4a+6b&=12\end{aligned}

ANSWER: To make the coefficients of a the same, we will multiply the first equation by 2. Doing so, we get

4a-8b=-2

Then, we will subtract the second equation from this new equation we just obtained. This looks like

Thus, we get the equation -14b=-14. If we divide both sides by -14, we get

b=\dfrac{-14}{-14}=1

Then, substituting b=1 into our very first equation, we get

2a-4=-1

Add 4 to both sides:

2a=3

Then, divide both sides by 2 to get

a=\dfrac{3}{2}

We have found both a and b, so we’re done.

NOTE: if your first step was to divide the second equation given in the question by 2, then that is perfectly valid. If you got the right answer, then that method (or any other way of multiply/dividing equations to make the elimination step work) is worth full marks.

FA15 – Proof

QUESTION: Prove that m(m+4)+3(2m+8) is equivalent to (m+4)(m+6).

ANSWER: We are going to expand the first expression, and the aim to factorise it to end up with the second expression. So, we get

m(m+4)+3(2m+8)=m^2+4m+6m+24 = m^2+10m+24

Then, we want two numbers that multiply to make 24 but add to make 10. Indeed, 6 and 4 satisfy these criteria, and so we get

m^2+10m+24=(m+4)(m+6)

So, we have shown that they are equivalent, and so we are done.

More Revision Cards answers

Open Search

Search for Revision Card Answers