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Reference Sheet: Programme F.5 — Linear Equations

Rules, procedures and drill, in the order of the book's programme. Every rule links to the unit in the lesson that derives it — if a rule here looks arbitrary, the link is where it stops being arbitrary.

1. Linear equations

From Unit 01.

A linear equation involves no power of the unknown (or unknowns) higher than the first. A linear equation in a single unknown is also called a simple equation. Each side of a simple equation, drawn against x, is a straight line; the solution is where the two lines meet.

2. Solution of simple equations

From Unit 01.

Simplify each side (expand brackets, collect like terms) to reach

ax+b=cx+d.

Take the same things from both sides to isolate the unknown: ax-cx=d-b, so

x=\frac{d-b}{a-c}\qquad\text{provided }a\ne c.

Examples.

9x-4=5x+12:\quad 4x=16,\quad x=4
3-2x=5x+17:\quad -14=7x,\quad x=-2\qquad\left(\frac{17-3}{-2-5}=-2\right)
4x+7-2x+1=3x+11:\quad 2x+8=3x+11,\quad x=-3

3. Checking by substitution

From Unit 01.

Substitute the answer into the original equation and evaluate the left and right sides separately. They must be equal.

Example. 3-2x=5x+17 with x=-2: left side 3+4=7; right side -10+17=7.

4. Equations that simplify to simple equations

From Unit 01.

Products of brackets can give x^2 terms. When both sides have the same x^2 term, subtracting it from both sides leaves a simple equation.

Examples.

(2x+1)(x-3)=(2x-1)(x+2):\quad 2x^2-5x-3=2x^2+3x-2,\quad -8x=1,\quad x=-\frac18
(4x+3)(3x-1)-(5x-3)(x+2)=(7x+9)(x-3):\quad 7x^2-2x+3=7x^2-12x-27,\quad 10x=-30,\quad x=-3

5. Beyond the book: when a=c

From Unit 01.

If the x terms cancel completely, the two lines are parallel.

After cancelling Lines Solutions
a false statement, e.g. 5=-1 parallel, apart none
a true statement, e.g. 6=6 the same line every x (an identity, \equiv)

Examples. 3x+5=3x-1 has no solution. 2(x+3)=2x+6 holds for every x: 2(x+3)\equiv2x+6.

6. Algebraic fractions: a numeric LCM

From Unit 02.

Multiply every term on both sides by the LCM (lowest common multiple) of the denominators, cancel, and solve the simple equation left. Keep each numerator in brackets: a minus sign in front of a fraction applies to its whole numerator.

Examples.

\frac{x-1}2+\frac{x+3}5=\frac{x+7}4\quad(\times20):\quad 10(x-1)+4(x+3)=5(x+7),\quad 14x+2=5x+35,\quad x=\frac{11}3
\frac x3-\frac{x-4}6=2\quad(\times6):\quad 2x-(x-4)=12,\quad x+4=12,\quad x=8

7. Algebraic fractions: an algebraic LCM

From Unit 02.

When the denominators contain x, the LCM is the product of the different factors. After multiplying out, the x^2 terms usually cancel (§4).

Examples.

\frac5{x+1}+\frac1{x-2}=\frac6x\quad\bigl(\times x(x+1)(x-2)\bigr):\quad 5x(x-2)+x(x+1)=6(x+1)(x-2)
6x^2-9x=6x^2-6x-12,\quad -3x=-12,\quad x=4.

Check: \frac55+\frac12=1.5 and \frac64=1.5.

\frac3{x-2}+\frac5{x-3}-\frac8{x+3}=0:\quad 3(x^2-9)+5(x-2)(x+3)-8(x-2)(x-3)=0,\quad 45x-105=0,\quad x=\frac73

8. Beyond the book: excluded values

From Unit 02.

Before multiplying, list every value that makes a denominator zero. An answer equal to one of them is not a solution; if it is the only answer, the equation has no solution.

Example. \frac{2x}{x-3}=\frac6{x-3}+1: exclude x=3. Multiplying by x-3: 2x=6+x-3, so x=3 — excluded. No solution.

9. Simultaneous linear equations with two unknowns

From Unit 03; the graph of an equation as all its solutions is F.4 Unit 01.

One linear equation in x and y has infinitely many solutions: a whole line. Two such equations may have just one pair of values satisfying both simultaneously: the point where their lines cross.

Example. y-x=3 is satisfied by (0,3), (1,4), (-2,1), …; together with x+y=7 only (2,5) fits both.

10. Solution by substitution

From Unit 03.

  1. Solve one equation for one unknown.
  2. Substitute that expression into the other equation and solve.
  3. Substitute back to find the second unknown.
  4. Check both values in both original equations.

Examples.

2x+y=11, 5x-3y=0: y=11-2x; 5x-3(11-2x)=0, 11x=33, x=3, y=5.

2x+3y=1, 5x-2y=12: y=\frac{1-2x}3; 5x-\frac{2(1-2x)}3=12, so 15x-2+4x=36, 19x=38, x=2, y=-1. Check: 4-3=1 and 10+2=12.

11. Solution by elimination

From Unit 03.

Multiply each equation by the coefficient of the chosen unknown in the other equation, so its two terms have the same size. Combine the equations to remove it (§12), solve for the other unknown, substitute into either original, and check in the other.

Scaling an equation does not change its line; combining two equations gives a line through the same crossing. Elimination picks the combination whose line is vertical or horizontal.

Example. 4x+3y=18 (1), 5x-2y=11 (2).

(1)\times2:\ 8x+6y=36\qquad (2)\times3:\ 15x-6y=33

Add: 23x=69, x=3. In (1): 12+3y=18, y=2. Check in (2): 15-4=11.

12. Add or subtract?

From Unit 03.

The equal-sized terms have Then
opposite signs (+6y, -6y) add the equations
the same sign (+6x, +6x) subtract one from the other

Example. 3x+5y=21 (1), 2x+3y=13 (2). Eliminate x: (1)\times2: 6x+10y=42; (2)\times3: 6x+9y=39. Same sign, so subtract: y=3. In (2): 2x+9=13, x=2. Check in (1): 6+15=21.

13. Substitution or elimination?

From Unit 03.

Both find the same crossing: substitution slides along one line to the other; elimination swings a line about the crossing. Use substitution when an unknown already has coefficient 1 or -1 (as y in 2x+y=11); otherwise elimination avoids fractions.

14. Beyond the book: no solution or infinitely many

From Unit 03.

If elimination removes both unknowns:

Result Lines Solutions
false, e.g. 0=1 parallel none
0=0 the same line infinitely many

Examples. 2x-y=3 and 4x-2y=5: (1)\times2-(2) gives 0=1, no solution. 2x-y=3 and 4x-2y=6: 0=0, every point of the line.

15. Simultaneous linear equations with three unknowns

From Unit 04.

Each equation is a plane; the solution is the point all three share.

  1. From one pair, eliminate one unknown: equation (4).
  2. From a different pair, eliminate the same unknown: equation (5). (An original that already lacks that unknown can be used as it stands.)
  3. Solve (4) and (5) as a two-unknown pair.

The book also calls this solving by equating coefficients.

Example. 2x+y-z=3 (1), x-y+2z=5 (2), 3x+2y+z=10 (3). Eliminate y:

(1)+(2): 3x+z=8 (4). (1)\times2-(3): x-3z=-4 (5).

(4)\times3+(5): 10x=20, x=2; in (4): z=2.

16. Three unknowns: the third value and the check

From Unit 04.

  1. Substitute the two values into one original equation to find the third unknown.
  2. Check all three values in the other two original equations.

Examples. Continuing §15: in (1), 4+y-2=3, so y=1. Check in (2): 2-1+4=5; in (3): 6+2+2=10.

5x-3y-2z=31 (1), 2x+6y+3z=4 (2), 4x+2y-z=30 (3): (1)\times3+(2)\times2 gives 19x+3y=101 (4); (3)\times2-(1) gives 3x+7y=29 (5). Then (4)\times7-(5)\times3: 124x=620, x=5, and y=2. In (3): 20+4-z=30, z=-6. Check in (1): 25-6+12=31; in (2): 10+12-18=4.

17. Pre-simplification: brackets

From Unit 05.

Expand and collect each equation until it has the form ax+by=c. Tidying leaves each line where it is. Answers need not be whole numbers.

Example. 3(x+y)-2(x-y)=16 and 2(x+3y)+(x-y)=21 give x+5y=16 (3) and 3x+5y=21 (4). (4)-(3): 2x=5, x=2.5; in (3): 5y=13.5, y=2.7. Check in (4): 7.5+13.5=21.

18. Pre-simplification: fractions

From Unit 05.

Multiply each equation by its own LCM, every term on both sides; then solve the tidy pair.

Example. \frac{x+1}3+\frac y2=3 (\times6): 2(x+1)+3y=18, so 2x+3y=16 (3). \frac x4-\frac{y-1}3=\frac{11}{12} (\times12): 3x-4(y-1)=11, so 3x-4y=7 (4). (3)\times4+(4)\times3: 17x=85, x=5; in (3): y=2.

19. Beyond the book: new variables

From Unit 05.

If the unknowns appear only inside the same expressions, such as \frac1x and \frac1y, name those expressions as new unknowns. The system is linear in them. (Further problems F.5 use u=\frac1{x+8y} and v=\frac1{8x-y}.)

Example. \frac3x-\frac2y=1, \frac6x+\frac4y=4. With u=\frac1x, v=\frac1y: 3u-2v=1 and 6u+4v=4. (1)\times2+(2): 12u=6, u=\frac12, so x=2; 2v=3u-1=\frac12, v=\frac14, so y=4.

Traps

  • Moving a term without changing its sign. 10+0.05m=4+0.08m is not 10-4=0.08m+0.05m (that gives m=46.2, where the bills are £12.31 and £7.69). Subtract 0.05m from both sides: 10-4=0.08m-0.05m, m=200.
  • Dividing by a-c when it is zero. 3x+5=3x-1 has no solution; it does not give x=\frac{-6}0.
  • Missing a term when clearing fractions. Every term on both sides is multiplied by the LCM, whole numbers included: \frac x3-\frac{x-4}6=2 becomes 2x-(x-4)=12, not 2x-(x-4)=2.
  • The minus in front of a fraction. -\frac{x+5}3\times12=-4(x+5)=-4x-20, not -4x+20.
  • Ignoring excluded values. \frac{2x}{x-3}=\frac6{x-3}+1 "gives" x=3, which makes the denominators zero: no solution.
  • Adding when the signs are the same. 6x+10y=42 and 6x+9y=39: adding gives 12x+19y=81 and removes nothing. Same signs, subtract.
  • Scaling only part of an equation. (2)\times2 means every term: 4x-3y=10 becomes 8x-6y=20, not 8x-6y=10. Scaling one side moves the line.
  • Eliminating different unknowns from the two pairs. One result is in x,y, the other in x,z: they cannot be solved together. Eliminate the same unknown twice.
  • Checking only where you substituted. The equation used to find the last unknown will always fit; check in the others.
  • Stopping at one unknown. Every unknown needs a value; substitute back.

Self-check

  1. Solve 7x-5=3x+11.
  2. Solve 3(2x-1)-2(x+4)=5(x-2).
  3. Solve (2x+3)(x-4)-(x-2)(x+5)=(x+1)(x-3).
  4. One plumber charges £40 call-out plus £30 an hour; another charges £25 plus £36 an hour. For what length of job do they charge the same, and how much?
  5. Solve \frac{x+1}4-\frac{x-2}6=\frac{2x+1}3.
  6. Solve \frac3{x-1}-\frac2{x+2}=\frac1x.
  7. Solve \frac{x+3}{x-1}=\frac4{x-1}+2.
  8. Solve by substitution: 3x+2y=7, 5x-3y=37.
  9. Solve by elimination: 5x+3y=29, 2x-7y=-13.
  10. Solve x+2y-z=-3, 2x-y+3z=14, 3x+y-2z=-1.
  11. Simplify and solve: 3(x-y)+2(x+2y)=23, 4(x+y)-(x-3y)=25.
  12. Solve \frac2x+\frac3y=2, \frac4x-\frac3y=1.

Fully worked solutions

1. Take 3x and add 5 on both sides: 4x=16, so x=4. Check: 28-5=23 and 12+11=23.

2. Left side 6x-3-2x-8=4x-11; right side 5x-10. So 4x-11=5x-10, -1=x. Check: 3(-3)-2(3)=-15 and 5(-3)=-15.

3. Left side (2x^2-5x-12)-(x^2+3x-10)=x^2-8x-2; right side x^2-2x-3. Subtract x^2 from both sides: -8x-2=-2x-3, so -6x=-1 and x=\frac16.

4. 40+30h=25+36h: 15=6h, h=2.5 hours. Both charge 40+75=25+90=£115. (Here a=30, b=40, c=36, d=25: \frac{25-40}{30-36}=2.5.)

5. LCM 12: 3(x+1)-2(x-2)=4(2x+1), so 3x+3-2x+4=8x+4, x+7=8x+4, 3=7x, x=\frac37.

6. Exclude x=0, 1, -2. Multiply by x(x-1)(x+2): 3x(x+2)-2x(x-1)=(x-1)(x+2), so x^2+8x=x^2+x-2, 7x=-2, x=-\frac27 (not excluded).

7. Exclude x=1. Multiply by x-1: x+3=4+2(x-1)=2x+2, so x=1 — excluded. There is no solution.

8. From the first, y=\frac{7-3x}2. Into the second: 5x-\frac{3(7-3x)}2=37; times 2: 10x-21+9x=74, 19x=95, x=5. Then y=\frac{7-15}2=-4. Check: 15-8=7 and 25+12=37.

9. Eliminate y: first \times7: 35x+21y=203; second \times3: 6x-21y=-39. Opposite signs, so add: 41x=164, x=4. In the first: 20+3y=29, y=3. Check in the second: 8-21=-13.

10. Eliminate z. (1)\times3+(2): 5x+5y=5, so x+y=1 (4). (1)\times2-(3): -x+3y=-5 (5). (4)+(5): 4y=-4, y=-1; then x=2. In (1): 2-2-z=-3, z=3. Check in (2): 4+1+9=14; in (3): 6-1-6=-1.

11. Expand: 5x+y=23 (3) and 3x+7y=25 (4). (3)\times7-(4): 32x=136, x=\frac{17}4=4.25. In (3): y=23-21.25=1.75=\frac74. Check in (4): 12.75+12.25=25.

12. With u=\frac1x, v=\frac1y: 2u+3v=2 and 4u-3v=1. Add: 6u=3, u=\frac12, so x=2. Then 3v=2-1=1, v=\frac13, so y=3. Check: 1+1=2 and 2-1=1.