# Equations and Inequalities

### Overview

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

3. Solve

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.

Exam­ple Question:

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

x =2

Answer y = 6 and x = 2

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

3.

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

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