Solving Linear Systems: Notes and Assignment

The Objective:

To find the number values of the VARIABLES (usually x and y) that satisfy all equations in the system. Substituting the solution values for the variables in the equations makes them both true statements.

Geometrically -- the solution is the point(s) of intersection of the lines in the system.

How Many Solutions? 3 Possiblities:

1) Skew Lines: meet at a single point -- 1 solution or root ( x, y )
look for different slopes using the coefficients of x and y
2) Parallel Lines: never meet -- no solutions
look for the same slopes but different y-intercepts (different constant term)
3) Coincident Lines: meet at every point -- infinite solutions.
look for the same slopes and same y-intercepts -- equations are identical.

Note: for coincident lines, the solution set is shown as indefinite coordinates (x, y) with y replaced with an expression in x. For example, if both lines are y = x + 1, the solution set is all points of the form (x, x + 1), where x is any real number. If both lines are y = – 2x + 3, the solution set is all points of the form (x, 2x + 3), where x is any real number.

4 Ways to Solve a Linear System

There are 4 methods for solving a system of linear equations in 2 variables. They are:

1) Graphically
2) Comparison Method
3) Substitution Method
4) Elimination Method

Which method we use depends on 2 things. First, how the question is worded; second, the form of the equations.

If the question says to solve graphically, then that's exactly what we do.
We use the slope and y-intercept or a table of values to graph the lines.
The solution is the coordinates of the point of intersection.

If the equations are given in standard form ( y = mx + b ), the Comparison method is best.
We set the expression for y in one equation = the expression for y in the other equation and solve.

If any of the equations has a variable with coefficient = 1 the Substitution Method is best.
We make a substitution expression for that variable and substitute it into the other equation.

If the equations are in the form Ax + By = C, the Elimination Method might be best.
We get the same coefficient for one variable in both equations, then eliminate it by adding or subtracting the equations.

Solving Graphically

In the graphic above, we solved the system with equations y = x + 1 and y = – x + 5.
Using the y-intercept and slope, we plotted the 2 lines and found their point of intersection is (2, 3). The solution is x = 2 and y = 3. If the equations are not in y = mx + b form, we put them in it. If the equations are such that we get ugly fractions, we use a table of values for x and y to plot points on the lines.

Comparison Method

If the equations are given in standard form ( y = mx + b ), we compare y's -- set the y's equal to each other like this:

Since y = x + 1 and y = – x + 5 we can set x + 1 = – x + 5 to get 2x = 4 so x = 2.
Since y = x + 1 and x = 2, y = 2 + 1 = 3. The solution is (2, 3).

We don't always compare y's. As long as we have an expression for the same term, we can use the comparison method.

Example: Solve the system if 2xy = 1 and 2x – 3y = – 5

Here we can set up equivalences for 2x from both equations and set them equal to each other.

2x = y + 1 ..... and ..... 2x = 3y – 5, ..... so

y + 1 = 3y – 5 .....or .....6 = 2y, which makes y = 3

Now, 2x = y + 1 ..... so ..... 2x = 3 + 1, which makes x = 2

The solution is (2, 3).

The Comparison and Substitution Methods are almost the same but not always. With comparison method, we equate 2 expressions for the same variable term derived from both equations in the system. With substitution, we get a substitution expression for one of the variables from one of the equations and then we substitute that into the other equation to get 1 equation in 1 variable.

Substitution Method

Here's the same example done by substitution instead of comparison. We use the first equation to get a substitution expression for y and then we plug that into the 2nd equation to find x.

2xy = 1 becomes y = 2x – 1

When we substitute, ..... 2x – 3y = – 5 ..... becomes ..... 2x – 3( 2x – 1 ) = – 5.

So 2x – 6x + 3 = – 5 .....or ..... – 4x = – 8 ..... so x = 2

Now we substitute x = 2 into y = 2x – 1 to get y = 3.

The solution is (2, 3).

Elimination Method

With this method, we eliminate one of the variables from the system by combining the 2 equations into one. We first arrange the equations so that one of the variables has the same coefficient in both equations. If that coefficient has opposite signs we add the equations together. If the signs are the same, we subtract one equation from the other.

Example: Solve this system using the elimination method: 2x + y = 7 and 3xy = 3

Solution: Since the coefficient of y is 1 in both equations and they have opposite signs, we add the 2 equations together to get:

5x + 0y = 10 which is really just 5x = 10, so x = 2.

Now we substitute x = 2 into either of the original equations to get y = 3.

In that example, the y's were already set up for us. Now one where we have to change one of the equations so that we can eliminate a variable.

Example: Solve this system using the elimination method: 5x + 2y = 13 and 3x + y = 7

Solution: We want the coefficient of y to be 2 in both equations, with one of them positive and the other negative, so that when we add the equations, the y terms cancel each other out. Let's multiply the 2nd equation through by – 2 and we'll have it.

– 2(3x + y = 7) = – 6x – 2y = – 14

Now when we add this to 5x + 2y = 13, we get – x + 0y = – 1, so x = 1.
When we substitute x = 1 into 3x + y = 7, we get y = 4.
The solution is (1, 4).

Now an example where we have to change both original equations to get the right coefficients so that we can eliminate a variable.

Example: Solve this system using the elimination method: 10a – 3b = – 7 and 4a + 5b = 22

Solution: If the coefficient of b in both equations was 15, we could eliminate the b terms. We multiply equation 1 by 5 and equation 2 by 3, to make the coefficients of b equal to – 15 and +15.

5(10a – 3b = – 7) [ 50a – 15b = – 35, and
3(4a + 5b = 22) [ 12a + 15b = 66

we add these equations to get
62a + 0b = 31, then we divide by 62, to get a = ½
Now we substitute a = ½ into 10a – 3b = – 7 to get b = 4.

To eliminate the a terms from these equations, we would've multiplied the 1st equation by 2 to get +20a and the 2nd equation by – 5 to get – 20a. Then we would add the equations to eliminate the a terms and solve for b.

Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.

Solving Linear Systems Assignment

1) State whether these lines are parallel, coincident or skew and how many solutions there are.

 a) . y = 2x – 5 . y = 2x + 1 b) . y = 3x + 4 . y = 2x – 4 c) . 3x + 6y = 12 . x + 2y = 4 d) . x – y + 8 = 0 2x – 2y – 10 = 0 e) 2x + y = 6 3x – y = 4

2) Solve these systems graphically:

 a) y = – 4x + 3 .... y = 2x – 3 b) y = 3x – 2 .... y = x + 2 c) y = – x – 4 .... y = x – 8

3) Solve the systems in question 2 by comparison.

4) Solve these systems by substitution.

 a) 4x + y = 3 2x – 3y = 12 b) 3x – 5y = 35 2.5x + y = 8.5 c) . x – 2y = 4 . 5x + y = 9 d) . 3x + 2y = 5 . 3x + 7y = 10

5) Solve the systems in question 4 by elimination.

Solutions

1) State which of the following systems of lines are parallel, coincident or skew and state how many solutions each system has.

 a) // lines no solution b) skew lines 1-solution c) coincident lines infinite solutions d) // lines no solution e) skew lines 1-solution

2) Solve these systems graphically:

3) Solve the systems in question 2 by comparison.

 a) set y = y – 4x + 3 = 2x – 3 so – 6x = – 6 and x = 1 since y = 2x – 3, .... y = 1solution: (1, – 1) b) set y = y 3x – 2 = x + 2 so 2x = 4 and x = 2 since y = x + 2, .... y = 4solution: (2, 4) c) set y = y – x – 4 = x – 8 so 4 = 2x which makes x = 2 since y = – x – 4, .... y = – 6solution: (2, – 6)

4) Solve these systems by substitution.

 a) 4x + y = 3 2x – 3y = 12substitution for y. . y = 3 – 4x2x – 3(3 – 4x) = 12 14x = 21 so x = 1.5 set x = 1.5, y = – 3solution: (1.5, – 3) b) 3x - 5y = 35 2.5x + y = 8.5substitution for y. . y = 8.5 – 2.5x3x – 5(8.5 – 2.5x) = 35 15.5x = 77.5 so x = 5 set x = 5, y = – 4solution: ( 5, – 4) c) . x – 2y = 4 . 5x + y = 9substitution for x. . x = 2y + 4. 5(2y + 4) + y = 9 10y + 20 + y = 9 11y = – 11 so y = – 1 set y = – 1, x = 2solution: (2, – 1) d) . .3x + 2y = 5 . .3x + 7y = 10substitution for 3x. . 3x = 5 – 2y (5 – 2y) + 7y = 10 5 + 5y = 10 . y = 1 and x = 1solution: ( 1, 1)

5) Solve the systems in question 4 by elimination.

 a) 4x + y = 3 multiply by 3 to get... 12x + 3y = 9 2x – 3y = 12 remains the same .....2x – 3y = 12add the equations to get .... ...........14x = 21so x = 1.5 and y = – 3 solution: (1.5, – 3) b) 3x – 5y = 35 remains same .... 3x – 5y = 35 2.5x + y = 8.5 times by 5 ....12.5x + 5y = 42.5add the equations to get .......15.5x = 77.5so x = 5 and y = – 4 solution: ( 5, – 4) c) . x – 2y = 4 remains same .... x – 2y = 4 . 5x + y = 9 times by 2 .... ..10x + 2y = 18add the equations to get .....11x = 22so x = 2 and y = – 1 solution: ( 2, – 1) d) . 3x + 2y = 5 remains same ....3x + 2y = 5 . 3x + 7y = 10 times – 1 ... – 3x – 7y = – 10add the equations to get ..... – 5y = – 5so x = 1 and y = 1 solution: ( 1, 1)

Note: in (d) above, we actually subtracted the 2nd equation from the first. Multiplying by – 1 and adding is exactly the same as subtracting.