Consistency of linear equations in two and three variablesGROUP 1

What is a linear equation A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.

Linear equations can have one or more variables. Linear equations occur abundantly in most subareas of mathematics and especially in applied mathematics. While they arise quite naturally when modeling many phenomena, they are particularly useful since many non-linear equations may be reduced to linear equations by assuming that quantities of interest vary to only a small extent from some "background" state. Linear equations do not include exponents.

What is a solution to an equation

In mathematics, to solve an equation is to find what values fulfill a condition stated in the form of an equation. These expressions contain one or more unknowns, which are free variables for which values are sought that cause the condition to be fulfilled. To be precise, what is sought are often not necessarily actual values, but, more in general, mathematical expressions. A solution of the equation is an assignment of expressions to the unknowns that satisfies the equation; in other words, expressions such that, when they are substituted for the unknowns, the equation becomes an identity.

With two equations and two variables, the graphs are lines and the solution (if there is one) is where the lines intersect. Let’s look at different possibilities.

11 12 1

21 22 2

a x a y b

a x a y b

Case 1: The lines intersect at a point

(x, y),the solution.

Case 2: The lines coincide and there are infinitely many solutions (all points

on the line).

Case 3: The lines are parallel so there

is no solution.

dependent

consistent consistent

independent

inconsistent

If we have two equations and variables and we want to solve them, graphing would not be very accurate so we will solve algebraically for exact solutions. We'll look at two methods. The first is solving by substitution.

The idea is to solve for one of the variables in one of the equations and substitute it in for that variable in the other equation.

Let's solve for y in the second equation. You can pick either variable and either equation, but go for the easiest one (getting the y alone in the second equation will not involve fractions).

Substitute this for y in the first equation.4 3y x

2 3 1

4 3

x y

x y

2 3 4 3 1x x Now we only have the x variable and we solve for it.

14 8x 4

7x Substitute this for x in one of the equations to find y.

Easiest here since we already solved for y.

4 54 3

7 7y

2 3 1

4 3

x y

x y

So our solution is 4 5

,7 7

This means that the two lines intersect at this point. Let's look at the graph.

x

y

4 5,

7 7

We can check this by subbing in these values for x and y. They should make each equation true.

4 52 3 1

7 7

4 54 3

7 7

8 151

7 716 5

37 7

Yes! Both equations are satisfied.

Now let's look at the second method, called the method of elimination.

The idea is to multiply one or both equations by a constant (or constants) so that when you add the two equations together, one of the variables is eliminated.

Let's multiply the bottom equation by 3. This way we can eliminate y's. (we could instead have multiplied the top equation by -2 and eliminated the x's)

Add first equation to this.12 3 9x y The y's are eliminated.

Substitute this for x in one of the equations to find y.

3 3

2 3 1x y

14 8x 4

4 37

y

2 3 1

4 3

x y

x y

4

7x

5

7y

So we arrived at the same answer with either method, but which method should you use?

It depends on the problem. If substitution would involve messy fractions, it is generally easier to use the elimination method. However, if one variable is already or easily solved for, substitution is generally quicker.

With either method, we may end up with a surprise. Let's see what this means.

3 6 15

2 5

x y

x y

Let's multiply the second equation by 3 and add to the first equation to eliminate the x's.3 3

3 6 15

3 6 15

x y

x y

0 0

Everything ended up eliminated. This tells us the equations are dependent. This is Case 2 where the lines coincide and all points on the line are solutions.

Now to get a solution, you chose any real number for y and x depends on that choice.

2 5x y

If y is 0, x is -5 so the point (-5, 0) is a solution to both equations.

3 6 15

2 5

x y

x y

If y is 2, x is -1 so the point (-1, 2) is a solution to both equations.

So we list the solution as:

2 5 where is any real numberx y x

Let's solve the second equation for x. (Solving for x in either equation will give the same result)

x

y

Any point on this line is a solution

Let's try another one: 2 5 1

4 10 5

x y

x y

Let's multiply the first equation by 2 and add to the second to eliminate the x's.

2 2

4 10 2x y

0 7

This time the y's were eliminated too but the constants were not. We get a false statement. This tells us the system of equations is inconsistent and there is not an x and y that make both equations true. This is Case 3, no solution.

x

y

There are no points of intersection

3333231

2232221

1131211

bzayaxa

bzayaxa

bzayaxa

The solution will be one of three cases:

1. Exactly one solution, an ordered triple (x, y, z) 2. A dependent system with infinitely many solutions3. No solution

Three Equations Containing Three Variables

As before, the first two cases are called consistent since there are solutions. The last case is called inconsistent.

Planes intersect at a point: consistent with one solution

With two equations and two variables, the graphs were lines and the solution (if there was one) was where the lines intersected. Graphs of three variable equations are planes. Let’s look at different possibilities. Remember the solution would be where all three planes all intersect.

Planes intersect in a line: consistent system called dependent with an infinite number of solutions

Three parallel planes: no intersection so system called inconsistent with no solution

No common intersection of all three planes: inconsistent with no solution

Steps to solving a system of equations in 3 Variables

Ensure that the equations are in standard form:

Ax + By + Cz = D

Remove any decimals or fractions from the equations.

Eliminate one of the variables using two of the three equations. Result will be a new equation with two variables.

Eliminate the same variable using another set of two equations. Result will be a second equation in two variables.

Solve the new system of two equations.

Using the solution for the two variables, substitute the values into one of the original equations to solve for the third variable.

Check the solution set in the remaining two original equations.

Example system of three equations

Example: solve for x, y and z

First, we need to ensure that all equations are in standard form, i.e. all variables are on the left side of the equation. Note that equation #3 is not in standard form. We need to get the y and z terms to the left side of the equation.

Put equations in Standard FormRewriting Equation 3 into standard form:

Add y to both sizes

Add 6z to both sides

Our 3 equations are now:

But, note that equations #1 contains a fraction, and #2 contains decimals. If we eliminate the fractions and decimals, the equations will be easier to work with.

Remove Fractions and Decimals We can eliminate the fraction in #1 by multiplying both sides of the equation by 4.

This is our new version of #1

• We can eliminate the decimals in #2 by multiplying both sides of the equation by 100.

• So, Let’s divide both sides by 5, so we have smaller numbers to work with.

This becomes our new #2

Note that all terms have a common factor of 5

Eliminate One VariableWe have rewritten our three equations as follows:

Now, to solve, we need to first get two equations with two variables. To do so, we eliminate one of the variables from two of the equations. Then eliminate the same variable from another set of two equations.

In the first column, we eliminated y from equations 1 & 2, resulting in equation A.

In the second column, we eliminated y from equations 2 & 3, resulting in equation B.

Solve the system in two variables

To solve our new system of equations A & B, the first step is to eliminate one of the variables

If we choose to eliminate x by addition, we must get the equations in a form such that the coefficients of x in the two equations are inverses of one another (for example: +1 and -1 or +5 and -5). To do so we need to multiply each of the equations by some factor.

Before we determine that factor, we need to determine what the resulting coefficient of x will be. It will need to be a common multiple of the current coefficients, 8 and 10. The least common multiple of 8 and 10 is 40, so we will want the coefficients of the x terms to be 40 and -40.

If we multiply equation A by 5 and equation B by 4, the resulting coefficient of x will be 40. However, we need one of those coefficients to be negative, so we’ll multiply the second equation by -4.

Equation A Equation B

Solve first for z, then for xAdd the new set of equations A and B to eliminate x, and then solve for z:

Now substitute the value for of z into equation A or B to solve for x; We are using equation B:

Add the two equations to eliminate x

Divide both sides by 102 and reduce

Substitute 1/6 for zSimplifySubtract 2 from both sidesDivide both sides by 10 and reduce

Use one of the original equations to solve for y

Next we need to substitute the values of x and z into one of the original equations to solve for y. We’ll use equation 2 (after decimals were removed):

Substitute the values for x and z

Simplify

Subtract 3 from both sizes

Multiply by (-1)

The solution for the system of equations is the ordered triple:

Matrix methods…..

Equations with infinite solutions

In the last row of the above augmented matrix, we have ended up with all zeros on both sides of the equations. This means that two of the planes formed by the equations in the system of equations are parallel, and thus the system of equations is said to have an infinite set of solutions. We solve for any of the set by assigning one variable in the remaining two equations and then solving for the other two.

Equations with no solutions

In the last row, we ended up with the equation 0 = 6 which we know can't be true and so we conclude that the system of equations has no solution.

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