This section builds on your Basic Algebra knowledge so that you can solve equations and inequalities. If you learn a few rules and practice a few questions its not too difficult. Some of the most demanding questions involving simultaneous equations (sound more difficult than they are) are Maths GCSE grade B.
Here are my notes. As usual to help me retain the knowledge I’ve devised some questions and these are followed by my approach and answers.
Equations and Inequalities Questions
1. Solve 8(p+3) — 2p = 36
2. Solve 4(3p — 5) = 20 + 4p
4. Solve 3(p + 7) ≥ 4(2p — 3)
5. Solve this pair of simultaneous equations.
b + a = 28
b + 2a = 36
6. Solve this pair of simultaneous equations.
3a + 6b = 24
5a + 17b = 50.5
Equations and Inequalities Approach
1. First expand the brackets, then combine common terms (in this case the “p’s”) and then solve the equation by adding, subtracting, multiplying or dividing both sides of the equation. Take as many steps as you wish. There is a temptation to combine a number of steps but there are no extra marks for doing this– so work at your level of competence.
2. When you have an unknown on both side of the equation, first multiply out any brackets and then add or subtract the unknowns from both sides of the equation so that the unknown is only on one side. This is definitely something that is easier to follow with a real example — see the answer below. This question gives a nice whole number as the solution but answers could be negative numbers, fractions or decimals.
3. First you need to multiply out the denominator (bottom of the equation). In this question you have to eliminate two different denominators on either side of the equation. To do this find their LCM (lowest common multiplier — see Factors, Powers and Roots). Multiply both sides of the equation by this LCM and then divide each side of the equation by the appropriate factor to eliminate the denominator.
Once you have eliminated the denominator(s) you can solve using the same methods as questions 1) and 2).
4. Inequalities are very much like equations to solve. A key point to remember is that when you are multiplying or dividing both side of an equality by a negative number you need to reverse the inequality sign. Just think of a very simple inequality 10 > 5. If you multiply both sides of the equation by –1 and left the sign unaltered you would have –10 > –5 which is incorrect, so correct by reversing the sign gives the correct answer; –10 < –5
5. Because simultaneous equations are true at the same time (that’s why they’re called simultaneous) you can add them together or subtract one from the other to solve them.
6. Sometimes you cannot simply add or subtract simultaneous equations to eliminate an unknown factor. You may have to multiply each equation by a different amount so that you can then eliminate one of the unknown factors.
3x + 2y = 18
5x + 7y = 52.
Here you would eliminate the unknown factor “x” by multiplying the first equation by 5 and and the second equation by 3 to give:
15x + 10y = 90
15x + 21y = 156
Then subtracting the first equation from the second gives:
11y = 66
y = 6 and substituting y for 6 in the first equation in the question gives:
3x + 12 = 18
3x = 6
Answer y = 6 and x = 2
Equations and Inequalities Answers
1. 8(p+3) — 2p = 36
8p + 24 — 2p = 36
6p + 24 = 36
6p = 36 — 24
6p = 12
p = 2
2. 4(3p — 5) = 20 + 4p
12p — 20 = 20 + 4p
12p = 40 + 4p
8p = 40
p = 5
4. 3(p + 7) > 4(2p — 3)
3p +21 > 8p — 12
3p + 33 > 8p
–5p > –33
when multiplying/dividing by negative number reverse the sign so:-
p < 33/5
5. b + a = 28
b + 2a = 36
subtracting one from the other gives a = 8 and b =20
6. 3a + 6b = 24
5a + 17b = 50.5
Multiply first equation by 5 and second equation by 3 so that we can eliminate the unknown factor “a”, gives:
15a + 30b = 120
15a + 51b = 151.5
subtracting one from the other gives:-
–21b = 31.5
–b = −(31.5÷21)
b = 1.5
substituting 1.5 for b in first equation gives:
3a + 9 = 24
3a = 15
a = 5
Answer a = 5 and b = 1.5
Below is not part of the answer but good to do if you have time. I find that I make a number of careless errors with these type of questions, I think I concentrate on the logic of solving the equations and rush the arithmetic.
Checking the second equation (substituting 5 for a and 1.5 for b) gives:
25 + 25.5 = 50.5