Linear equations

LINEAR EQUATIONS IN

TWO VARIABLES

Review video from last week’s lecture

System of equations A pair of linear equations in two variables

is said to form a system of simultaneous linear equations.

For example: 2x – 3y +4 = 0 and x + 7y -1 = 0

These two equations form a system of two linear equations in variables x and y.

Continue… The general form of a linear equation x and y is

Ax + by +c = 0, where a and b is not equal to zero and are real numbers.

A solution of such an equation is a pair of values. One is for x and the other for y. Once the values of x and y are represented the two sides of the equation hold for them to be equal.

Every linear equation in two variables has infinitely many solutions which can be represented on a certain line.

Graphical Solutions of a Linear Equation

Let us consider the following system of two simultaneous linear equations in two variable:

2x-y=-1 and 3x+2y=9 Here we assign any value to one of the

two variables and then determine the value of the other variable from the given solution by using a t-chart

T-chart

For the equation: 2x-y=-1 solve for y y=2x+1 Plug x=0,2 to get y values

3x+2y=9 solve for y y=9-3x Plug x=3,-1 to get y valuesAny questions on how to Complete t-chart?

x 0 2y 1 5

x 3 -1y 0 6

Graph equationsX Y

0 1

2 5

X Y

3 0

-1 6

Y=2x+1

Y=9-3x

-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.501234567

Y-Values

Algebraic methods of solving liner equations

The most common used algebraic methods of solving simultaneous linear equations in two variables are:

Method of by substitution. Method by equating the coefficient. Method by elimination

Method by Substitution Solve the equations given by solving for x and

substitute the x-value of (i) first equation into (ii) second equation to get one equation

(i) x+2y=-1 and (ii)2x-3y=12(i): x=-2y-1 plug this equation into (ii) by substituting value of x2(-2y-1)-3y=12 distribute and simplify-4y-2-3y=12 combine like terms and solve for y-7y-2=12-7y=14y=-2 putting the value of y in equation (i) w getx=-2y-1x=-2(-2)-1x=3 Hence solution of the equation is (3,-2)

Use Method by Substitution

Try on your own Use previous slide to solve for x and y in

these two equations given: y=5x-1 and 2y=3x+12

Solution y=5x-1 (i) and 2y=3x+12 (ii)Step 1: substitute (i) into (ii) from y value.Step 2: 2(5x-1)=3x+12 distributeStep 3: 10x-2=3x+12 combine like terms to solve for xStep 4: 7x-2=12

7x=14 x=2

Step 5: Use value of x to solve for y in (i) equationStep 6: y=5(2)-1 y=9Step 7: Solution: x=2 and y=9, check your work by

plugging in these values in both equations to make sure left equals right in both equations.

Any questions?

Method by Elimination In this method, we eliminate one of the

two variable which can easily be solved. Putting the value of this variable in any of

the given equations. The value of the other variable can be obtained.

For example we want to solve: 3x+2y=11 and 2x+3y=4

Continue… (i) 3x+2y=11 and (ii) 2x+3y=4 First we can use a method by equating the coefficient

which means multiplying (i) by 3 and (ii) by -2. We get 3(3x+2y=11)------9x+6y=33

-2(2x+3y=4)------- -4x-6y=-8Now we can eliminate y values by adding both (i) and

(ii) 9x+6y=33+ -4x-6y=-8----------------5x=25 solve for xx=5

Continue… Putting the value of y in equation (i) 3x+2y=11 3(5)+2y=11 15+2y=11 2y=-4 y=-2 Solution: x=5 and y=-2 Check solutions

Use Method by Elimination Try on your own Use previous slides to solve for x and y in

these two equations given: x+3y=-5 and 4x-y=6

Solution x+3y=-5 (i) and 4x-y=6 (ii) Step 1: Multiply (i) by -4 to eliminate x. Step 2: -4(x+3y=-5)------- -4x-12y=20 Step 3: Add both equations and combine like terms to

eliminate x. Step 4: 4x-y=6 + -4x-12y=20

-----------------------13y=26-y=2…..multiply by (-1)….. y=-2

Step 5: Plug in y=-2 into equation (i)x+3(-2)=-5…. x-6=-5 x=1Step 6: Check solutions

Review Given two equations, with two variables, we

are then able to use different methods to solve for x and y

Once we found the values of x and y, are we finished? We need to make sure both values hold for both

equations. What are the two methods we went over?

The methods are methods by substitution and method by elimination.

Any questions?