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### Grade 6 - Mathematics12.10 Linear Equations

Linear equations are equations with variables to 1th power. In other words, linear equations are equations with degree 1. Linear equations are mathematical statement that performs functions of addition, subtraction, multiplication, and division.

Examples:
x + 5 = 10
3x + 2 = 3x - 9
6a - 5 = 4
3a + 2b = 6
3n = 15
n/2 = 12

However, variable(s) in linear expressions

• cannot have exponents example: x2
• cannot multiply or divide each other example: xy or x/y
• cannot be found under a root sign or square root. example: sqrt(x)

The following are NOT linear equations:
x2+2x+5=0
3a3-3a2+2a-5=0
4xy + 2 = 12

Solving Linear Equations
 One Variable One Variable - Multiple terms Method: Bring the variable term to one side and numbers on the other side of the equation. Question: Solve w + 8 = 17 . The linear equation is read as: a number plus 8 equals 17. When you see a plus sign use the subtration operation. So, subtracting 8 both sides we get w + 8 - 8 = 17 - 8 w = 9 Question: Solve m - 11 = 8. This linear equation is read as: a number minus 11 equals 8. When you see a minus sign use the addition operation, therefore adding 11 both sides of the equation we get m - 11 + 11 = 8 + 11 m = 19 Question: Solve 4z = 36. Four times z equals 36. 4 x z = 36 dividing both sides by 4 4z / 4 = 36/4 z = 9 Question: Solve 2x = 10. Twice a number equals 10. 2x = 10 Dividing both sides by 2 2x/2 = 10/2 x = 5 Question: Solve (1/3)x = 12 x has a fractional coefficient, multiply each side by 3. 3(1/3)x = 12 x 3 x = 36 Question: Solve (-2/5)x = 24 multiply each side by reciprocal of -2/5 that is -5/2 -5/2 x (-2/5)x = -5/2 x 24 x = -60 Question: Solve x/3 = 5. Since x is divided by 3, multiply both sides of the equation by 3, 3(x/3) = 3 x 5 x = 15 Question: 9/3y = 3. Method 1: 3y/9 = 1/3 Multiply both sides by 9 9(3y/9) = 9(1/3) 3y = 9/3 3y = 3 y = 3/3 = 1 Method 2: 9/3y =3 3y is in the denominator, so multiply both sides by 3y, (9/3y) 3y = 3 . 3y 9 = 9y 9/9 = y y = 1 Question: Solve 1/n = 4 Method 1: 1/n = 4 multiply both sides by n n(1/n) = 4 x n 1 = 4n 4n = 1 dividing both sides by 4 4n/4 = 1/4 n = 1/4 Method 2: 1/n = 4 Inverse both sides n = 1/4 Method: Bring the variables to one side and numbers on the other side of the equation. Question:Solve 3x - 7 = 23 add 7 both sides 3x - 7 + 7 = 23 + 7 3x = 30 div each side by 3 3x/3 = 30/3 x = 10 Question:Solve 4x + 9 = 53 subtract 9 both sides 4x + 9 - 9 = 53 - 9 4x = 44 divide both sides by 4 4x/4 = 44/4 x = 11 Question:Solve (1/3)x + 2 = 7 subtracting 2 both sides (1/3)x + 2 - 2 = 7 - 2 (1/3)x = 15 multiplying both sides by 3 3(1/3)x = 15 x 3 x = 45 Question:Solve (1/5)x - 6 = 9 adding 6 both sides (1/5)x - 6 + 6 = 9 + 6 (1/5)x = 15 multiply each sides by 5 5(1/5)x = 15 x 5 x = 75 Question:Solve (2/5)x + 3 = 21 Subtracting 3 both sides (2/5)x +3 - 3 = 21 - 3 (2/5) x = 18 Multiplying both sides be the reciprocal of 2/5 that is 5/2 5/2(2/5) x = (5/2)18 x =45 Question:10x - 5 = 2x + 21, then x = ? Bring the variables on one side and numbers on the other Adding 5 both sides 10x - 5 + 5 = 2x + 21 - 5 10x = 2x + 16 Subtracting 2x both sides 10x - 2x = 2x + 16 - 2x 8x = 16 Dividing both sides be 8 8x/8 = 16/8 x = 2 Question:Solve 1/x + 5 + 3 = 10 + 1 subtracting 5 both sides 1/x + 5 - 5 + 3 = 10 + 1 - 5 1/x + 3 = 6 subtracting 3 both sides 1/x + 3 - 3 = 6 - 3 1/x = 3 x = 1/3 Question:Solve 10(x-4)=5(x-1)+5, then x = ? Using distributive law 10x - 40 = 5x - 5 + 5 10x - 40 = 5x Adding 40 both sides 10x - 40 + 40 = 5x + 40 10x = 5x + 40 Subtracting 5x both sides 10x - 5x = 5x + 40 - 5x 5x =40 Dividing both sides by 5 5x/5 =40/5 x = 8

Directions: Solve the following linear equations. Also write at least 10 examples of your own.
•  Q 1: 5n = 40, then n = ?867 Q 2: 8n - 5 = 123, then n = ?151614 Q 3: 6x - 5 = 2x + 9, then x = ?5/36/57/2 Q 4: 10(x-4)=4(2x-1)+5, then x = ?42/541/241/3 Q 5: 10x - 9 = 41, then x = ?645 Q 6: 2.5x + 50 = 80, then x = ?12610 Q 7: 3x - 5 = 2x + 3, then x = ?8-87 Q 8: z + 20 = 5z - 44, then z = ?161214 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!