Linear Equations

Objectives

•To solve linear equations in one unknown.

•To transpose and solve formulae.

•To construct linear equations.

•To solve simultaneous linear equations by substitution and elimination methods.

•To use linear equations to solve problems.

Revision of solving linear equations

• A linear equation has a variable whose value is unknown. The power of the variable is 1.

• For example 7x – 3 = 6.

• 4x2 + 5 = 12 is not linear. It is a quadratic equation because x has a power of 2.

• There are two important rules to remember when solving equations:

• Undo what has been done

• Do the same to both sides of the equations

• Two other ‘rules’:

• Write your x’s like two ‘c’s back-to back. Not like this: x

• Write the equation out and show working, keeping the ‘=‘ in a column.

When more than one operation is performed on the unknown we need to solve the equation in several steps.

We can do this by performing the same operations to both sides of the equals sign to keep the equation balanced.

For example,4x + 5 = 29

4x = 24subtract 5 from both sides:

– 5– 5

÷ 4÷ 4

divide both sides by 4: x = 6

Check that 4 × 6 + 5 is equal to 29 in the original equation.

Revision of solving linear equations

These equations can be solved by performing the same operations on both sides until the solution is found.

Check by substituting x = 0.5 into the expressions in the original equation. Both sides are equal to 2, so the solution is correct.

8x – 2 = 2x + 1

6x – 2 = 1

add 2 to both sides: + 2+ 2

÷ 6÷ 6divide both sides by 6:

x = 0.5

6x = 3

subtract 2x from both sides: – 2x– 2x

Unknown on both sides

Equations can contain brackets. For example,

2(3x – 5) = 4x

To solve this we can

multiply out the brackets: 6x –10 = 4x

+ 10 + 10

add 10 to both sides:

2x –10 = 0

- 4x - 4x

subtract 4x from both sides:

2x = 10÷ 2 ÷ 2

divide both sides by 2: x = 5

Equations with brackets

In this example the whole of one side of the equation is divided by 5.

2x + 75

= x – 1

To remove the 5 from the denominator we multiply both sides of the equation by 5.

2x + 7 = 5(x – 1)

swap sides: 5x – 5 = 2x + 7

add 5 to both sides:

3x – 5 = 7subtract 2x from both sides:

3x = 12

divide both sides by 3: x = 4

expand the brackets: 2x + 7 = 5x – 5

Equations containing fractions

When both sides of an equation are divided by a number we must remove these by multiplying both sides by the lowest common multiple of the two denominators. For example,

5x – 34

=2x – 1

3

What is the lowest common multiple of 4 and 3?

The lowest common multiple of 4 and 3 is 12. Multiplying every term by 12 gives us:

12(5x – 3)4

=12(2x – 1)

3

3

1

4

1

which simplifies to: 3(5x – 3) = 4(2x – 1)

Equations containing fractions

We can then solve the equation as usual.

3(5x – 3) = 4(2x – 1)

expand the brackets: 15x – 9 = 8x – 4

add 9 to both sides: 15x = 8x + 5

subtract 8x from both sides: 7x = 5

divide both sides by 7: x =57

Although this answer could be written as a rounded decimal, it is more exact left as a fraction.

Equations containing fractions

Revision of solving linear equations

• For further help, check out these Youtube clips.

• http://www.youtube.com/watch?v=W7fPsSw74TM&feature=related

Lesson 1 – basic equations

• http://www.youtube.com/watch?v=58mHEQR8GFs&feature=related

Lesson 2 – equations with fractions

• http://www.youtube.com/watch?v=juG-iIuTJQE&feature=related

Lesson 3 – harder equations with fractions

• http://www.youtube.com/watch?v=9teKXGoWIQM&

Equations with unknown on both sides

Constructing Linear Equations

Simultaneous Equations

• Find three solutions to the equation 2y + x = 14

Simultaneous Equations

• A linear equation with two unknowns has an infinite number of solutions.

• If we plotted them on a graph we would have a straight line.

Simultaneous Equations

• A linear equation with two unknowns has an infinite number of solutions.

• If we plotted them on a graph we would have a straight line.

• If we plotted two of these straight line graphs on the same set of axes, they would intersect at one point provided the lines are not parallel.

Simultaneous Equations

• A linear equation with two unknowns has an infinite number of solutions.

• If we plotted them on a graph we would have a straight line.

• If we plotted two of these straight line graphs on the same set of axes, they would intersect at one point provided the lines are not parallel.

• Hence, there is one pair of numbers which solve both equations simultaneously.

• We will look at two methods for solving

simultaneous equations.

Solving simultaneous equations by substitution

• Solve the equations 2x − y = 4 and x + 2y = −3.• Solution by substitution• 2x − y = 4 (1) First label your equations (1) and (2)• x + 2y = −3 (2)• Write one unknown from either equation in terms of the other unknown.• Rearranging equation (2) we get x = −3 − 2y.• Then substitute this expression into equation (1).• 2(−3 − 2y) − y = 4• −6 − 4y − y = 4• −5y = 10• y = −2• Substituting the value of y into (2) gives us x + 2(−2) = −3• x = 1• Check this answer solves both equations• This means that the point (1, –2) is the point of intersection of the graphs of the two

linear relations.

Solving simultaneous equations by elimination

• Solve the equations 2x − y = 4 and x + 2y = −3.• Solution by elimination• 2x − y = 4 (1) Again, label your equations (1) and (2)• x + 2y = −3 (2)• If the coefficient of one of the unknowns in the two equations is the same, we can

eliminate that unknown by subtracting one equation from the other. It may be necessary to multiply one of the equations by a constant to make the coefficients of x or y the same for the two equations.

• To eliminate x, multiply equation (2) by 2 and subtract the result from equation (1).• Equation (2) becomes 2x + 4y = −6. Call this new equation (2’)• Then 2x − y = 4 (1)• 2x + 4y = −6 (2’)• Subtracting (1) − (2’): −5y = 10• y = −2• Now substitute for y in equation (1) to find x, and check as in substitution method.

• For more help understanding the substitution method, try

• http://www.youtube.com/watch?v=8ockWpx2KKI

• For more help understanding the elimination method, try

• http://www.youtube.com/watch?v=XM7Q4Oj5OTc