BIRKBECK MATHS SUPPORT
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Answers to Algebra 1
Here we give the answers and some workings to the questions at the end of
the Algebra 1 worksheet. Answers are shown in blue
A) Starting simplifying
1) 2a + 3a = 5a
6) 2z –7z + 4y + 2y = –5z + 6y
2) 7a – 3a = 4a
7) a2 –3a2 + b – 4b = –2a2 – 3b
3) 8z + 1z + 5a = 9z + 5a
8) a2 –3a2 + a – 4a = –2a2 – 3a
4) –5a + 10a – 2z = 5a – 2z
9) b2 –3b2 + 4b – b = –2b2 +3b
5) – 2a – a – 8a = – 11a
10) z2 –3a2 + b – 4a = z2 –3a2 + b – 4a
Notice that 10) is a bit of a trick question as it cannot be simplified
B) 4a2 + 2b – 7a +3b + 8b – 9 (a – 2) + 18
1) List all the constants in the expression written above
4, 2, -7, 3, 8, -9, -2, 18
2) List all the variables in the expression written above
a, b, (note that a2 and b are expressions of the variables a and b)
3) Write the algebraic expression above in its simplest form
Step 1: multiplying out the bracket gives
4a2 + 2b – 7a +3b + 8b – 9a + 18+ 18
Step 2: re-ordering and simplifying
4a2 – 7a – 9a + 2b + 3b + 8b + 18+ 18 = 4a2 – 16a + 13b + 36
www.mathsupport.wordpress.com Jackie Grant, Birkbeck College, 2012
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C) Substitution
Substitute values into the expressions below to calculate the total value of the
expression: a = 7, b = -2 , y = 3, z = 1. Note you should get the same answers
if you substitute the values into your answers for part A).
1) 2a + 3a = 2 (7) + 3(7) = 14 + 21 = 35
2) 7a – 3a = 7(7) – 3(7) = 49 – 21 = 28
3) 8z + 1z + 5a = 8(1) + 1(1) + 5(7) = 8 + 1 + 35 = 44
4) –5a + 10a – 2z = –5(7)+ 10(7) – 2(1) = –35 + 70 –2 =33
5) – 2a – a – 8a = – 2(1) – 1 – 8(1) = – 2– 1 – 8 = – 11
6) 2z –7z + 4y + 2y = 2(1) –7(1) + 4(3) + 2(3) = 2 – 7 + 12 + 6 = 13
7) a2 –3a2 + b – 4b = (7)2 –3(7)2 + (–2) – 4(–2) = 49 – 3(49) – 2 + 8 = 92
8) a2 –3a2 + a – 4a = (7)2 –3(7)2 + 7 – 4(7)= 49 – 3(49) + 7 – 28 = –119
9) b2 –3b2 + 4b – b = (–2)2 –3(–2)2 + 4(–2) – (–2) = 4 – 3(4) – 8 + 4 = – 8
10) z2 –3a2 + b – 4a = (1)2 –3(7)2 –2 – 4(1) = 1 – 3(49) – 2 – 4 = – 152
D) Practicing simplifying the following expressions using the BIDMAS rules
1) 2 (z + 2) + 8 = 2 z + 4 + 8 = 2z +12
2) 3(z – 2) – 2(z + 3) =3z – 6 – 2z – 6 = z – 12
3) 3(a – 2) – 2(b – 3) = 3a – 6 – 2b + 6 = 3a – 2b
4) 2 (z2 – 2) + 2(z – 4) = 2 z2 – 4 + 2z – 8 = 2 z2 + 2z – 16
5)
6) 2 (y2 – 2) + (–2 y)2 + 4 = 2y2 – 4 + 4 y2 + 4 = 6y2
7)
=
8) 2 (z2 – 2) + (2z)2 + 4 = 2 z2 – 4 + 4z2 + 4 = 6z2
www.mathsupport.wordpress.com Jackie Grant, Birkbeck College, 2012
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9)
10)
=
=
11) 3a(b – a2) – a2(2 – 3a) = 3ab – 3a3 – 2a2 + 3a3 =3ab – 2a2
12)
E) Re-arrange each of these equations to make
the subject
(there are many ways to re-arrange these equations, these show one method)
1) b + 4 = a ; b = a – 4
2) 4b = y ;
3) 4b = y + a ;
4) 3b – y = 2a ;
5) 3b – a = 5a ;
6) 3(b – a) = 9a ;
7) 2 (b2 – 2) = 4 ; b2 – 2 = 2 ; b2 = 4 ; b =  2 (note: ‘’ means plus or minus)
8) 4 = b +2y ; b + 2y = 4 ; b = 2y – 4
9)
10)
; b = 4a
; 4 = 2ab ; 2ab = 4 ;
11)
12)
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F) Solve each of these equations to find
1) b + 2 = 8 ; b = 8 – 2 ; b = 6
2) 3b – 2 = 7; 3b = 5 ;
3) 3(b – 2) = 9 ; b – 2 = 3 ; b = 5
4)
5) – b + 2 = 8 ; 2 = 8 + b ; b + 8 = 2 ; b = 2 – 8 ; b = 6
6) b2 – 9 = 7 ; b2 = 7 + 9 ; b2 = 16 ; b =  4
7) – 3(2 – b) = 12 ;
8)
9) 3(b – 2) + 2b = 9 ; 3b – 6 + 2b = 9 ; 5b – 6 = 9 ; 5b = 15 ; b = 3
10)
11) – 3(– b – 2) = – 9 ; 3b + 6 = – 9 ; 3b = –15 ; b = –5
12)
www.mathsupport.wordpress.com Jackie Grant, Birkbeck College, 2012
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G) Some word questions
1) Using T to represent the price of a cup of tea and C to represent the price of
a cup of coffee write an equation to show that 2 cups of tea and 3 cups of
coffee cost £5.40
a) One cup of tea costs £1.20. Substitute this value into your equation and
simplify.
b) Solve this equation to find the price of one cup of coffee.
.
So the cost of a coffee is £1.00
2) The formula for calculating the time taken to cook a turkey is
where T is the time in minutes needed to cook a turkey, and p is the weight in
pounds.
a) How many minutes does it take to cook a 10 pound turkey
.
So the time to cook the turkey is 170 minutes.
b) Re-arrange the equation to make p the subject, and then calculate how
many pounds the turkey is if the cooking time is 4 hours.
www.mathsupport.wordpress.com Jackie Grant, Birkbeck College, 2012
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2) The total cost of borrowing money can be calculated using the formula
Where
is the number of years the money is borrowed for, is the interest rate
as a decimal,
is the amount of money borrowed and
is the total amount
that is repaid.
a) If the amount borrowed is £10,000, the interest rate is 5% (or 0.05 as a
decimal) and the number of years is 10. Calculate the total amount that is
repaid.
; note that we have used a calculator for this calculation.
So the total amount that is repaid is £16,289.
b) Re-arrange the equation to make the subject, and then calculate the
interest rate needed if the amount borrowed is £1,000, the number of years
is 3 and the total amount to be repaid is £10,000.
. This was a little tricky because of the indices (or powers) and
we have used
which means the same as
. If you want to review
indices visit the Number section at www.mathsupport.wordpress.com
Now substituting
and
we get
So the interest rate is 115%. Meaning that if you borrow £1,000 at an interest
rate of 115% over 3 years you end up repaying £10,000.
www.mathsupport.wordpress.com Jackie Grant, Birkbeck College, 2012
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