| solving systems of equations |
When we solve simultaneous equations or systems of equations in the same variables, we look for common solutions for all the equations in the system. It can be done graphically as well as algebraically, since linear equations, as their name suggests, when graphed in either 2 or 3 dimensions, produce a line.
The 3 linear equations depicted in this graph are y = x – 1, y = x + 3, and y = – x – 3.
Notice that the blue and red lines are parallel and therefore have no point of intersection. If you have already studied the analytic geometry of lines, you'll understand that this is because they have the same slope but different y-intercepts. The black line has a different slope and therefore intersects each of the 2 parallel lines at a single point. We will use algebraic methods to find the coordinates of these points when we solve the system of equations associated with these lines.
Before we do however, let's discuss the geometric interpretations of the solutions. When we solve two or more linear equations, we look for the points that the two or more lines have in common. There are three possibilities.
Because of these 3 possibilities, we must always check our solutions in all of the original equations in the system.
The 3 possibilities for a conic and a line
Some systems of equations are made up of a mix of linear and other types of equations. For instance, we may be trying to find the points of intersection of a line with a circle or parabola. In such cases we attempt to solve both linear and quadratic equations simultaneously and might find that there are no solutions, one solution, or two solutions, since a line and a circle or parabola can have one, two, or no points of intersection. We will discuss both types of approaches and formulate algebraic techniques for solving different types of simultaneous or systems of equations.
The 2 methods used to solve systems of equations are the elimination method, in which we eliminate one of the variables to solve for the other, and the substitution method in which we substitute an expression for one of the variables. There are text books which mention the comparison method -- which, in my opinion, is just a fancy-shmancy name for the substitution method so it will be covered in that section.
The main difference of course between these equations and anything we've covered up until now, is that there are 2 variables in a system of equations. That is because they represent algebraic relations between variables. If you have not yet worked with equations in 2 variables, see the lesson entitled relations and functions in another section of this site.
In many cases, the word problems associated with systems of equations could be solved using a single variable. This is accomplished by performing a substitution before creating the equations.
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For example:
The length of a rug is 3 metres longer than its width.
If the area of the rug is 108 square metres, what are its dimensions?
We could say let x = the length and let y = the width.
But, since we know that the length is 3 metres longer than the width,
if we set x = the width, we can then express the length as x + 3
So we could solve this question using a single variable.
Since we wish to learn how to solve systems in 2 variables, we will do all the questions in this section that way.
| intro | linear systems | mixed systems | examples | practice | solutions |
| linear equations are equations in two variables, both raised to the first power |
The two most common forms of the linear equation are:
the general form Ax + By + C = 0
and the slope/y-intercept form y = mx + b.
When we solve a system of linear equations, we have two choices of approach: the elimination method or the substitution method. Which one we choose depends on the type of equations we have. If the coefficients of the variables are such that substitution is simple, we should use that method. However, elimination is very useful when the equations are presented in General Form so that substitution requires extensive work with fractions.
For instance, say we have to find the solutions for x + 2y = 7 and x – y = –2.
Substituting x = y – 2 from the 2nd equation into the 1st will give me my solution with a minimum of calculation
However, if we have to solve 5x – 3y = 22
and 7x + 13y = 104, substitution would be difficult.
If we use the first equation to substitute for x, we get 
, then we multiply that by 7 when substituting into the 2nd equation.
This approach would be far from efficient. Elimination would be the way to go.
The Elimination Method
When using this method, we perform algebraic operations on the equations so that the absolute value of the coefficients of one of the variables is the same in two of the equations. Then we add or subtract the equations to eliminate that variable. This results in a single equation in one variable which we then solve. Substituting the value we found into one of the original equations will give us the solution for the other variable.
Here's an example:
Determine the common solution if any for
3x – 2y = – 12
2x + y = – 1
Multiply the 2nd equation by 2: 4x + 2y = – 2
Rewrite the 1st equation 3x – 2y = – 12
Now add them, eliminate y: 7x = – 14, so x = – 2.
When we substitute x = – 2 into the 1st original equation, we get y = 3.
Checking these two values in the 2nd equation verifies our solution.
When we graph the 2 lines in this system, they intersect at ( –2, 3 ).
Now a word problem solved with a system.
question:
A group of LCC students went to the greasy spoon for burgers and fries. 5 burgers and 3 fries cost $11.25. One burger and 3 fries cost $5.25.
How much would it cost for 7 burgers and 5 fries ?
solution:
Let b = the price of 1 burger, and let f = the price of 1 order of fries.
The 1st equation says: 5b + 3f = 11.25
The 2nd equation says: b + 3f = 5.25
Subtract them 4b = 6.00 which means that b = $1.50
Substitute b = $1.50 in 2nd equation to get f = $1.25.
So, 7 burgers and 5 fries will cost: 7( 1.50 ) + 5 ( 1.25 ) = $16.75
It would cost $16.75 for 7 burgers and 5 fries.
The Substitution Method
With the substitution method, we use one of the equations to generate a substitution expression for one of the variables.Then we substitute that expression for the variable into another equation in the system and solve.
Here's an example:
Determine the common solution if any for 3x + 2y = 19
and x – 3y = – 12
Use the 2nd equation to substitute x = 3y – 12 into the 1st equation.
Substituting into the 1st equation we get 3(3y – 12) + 2y = 19.
We remove brackets, collect terms to get 11y – 36 = 19 so 11y = 55.
When we solve this for y, we get y = 5
Substituting y = 5 into the 1st equation, we get x = 3.
Generally, it is best to use the substitution method when the coefficient of one of the variables in the system is ±1.
If there are more than two equations in two unknowns in the system, we solve two of them and then check that their common solution also satisfies the other equations in the system. When trying to solve systems of equations in more than two variables, it is best to use matrices.
If both equations are presented in the slope/y-intercept form of y = mx + b, we have 2 ready-made substitution expressions -- since both equations are statements about the value of y. Setting the two equations equal then is really just substituting for y in one of the equations. Some algebra books refer to this method as the comparison method -- because we seem to be comparing the y values to each other. Just look at this example and you'll understand why I call it substitution.
y = 2x + 1
y = x + 3
2x + 1 = x + 3
It's pretty obvious that all I did was replace the y in the 2nd equation with its substitution expression from the 1st equation. And since the y's are the same at the points of intersection, I can substitute one for the other. So why bother to call this method something other than substitution?
| intro | linear systems | mixed systems | examples | practice | solutions |
For systems of mixed equations such as linear and quadratic equations, we always use a substitution expression from the linear equation.
Here's an example:
Determine the common solutions if any for: x 2 + y 2 = 8 (circle)
and – 3x + y = 8 (line)
Use the linear equation to get y = 3x + 8
substitute it into the 1st equation. x2 + (3x + 8) 2 = 8
x 2 + 9x 2 + 48x + 64 = 8
10x 2 + 48x + 56 = 0.
We divide through by 2 and factor to get (5x + 14) (x + 2) = 0.
So, our two solutions are x = – 2, and x = 
.
We substitute these values into the 2nd equation we get y = 2 or y = 
. The coordinates of our two solutions are ( –2, 2 ) and 
. The equations in this question represent a circle and a line. Since we found two points of intersection, the line is secant to the circle at these two points. Had the line been tangent to the circle, we would have found only one solution. Had the line been disjoint to or remote from the circle, there would be no solution.
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| intro | linear systems | mixed systems | examples | practice | solutions |
| a) x + y = 6 becomes x = 6 – y x – y = 2 becomes (6 – y) – y = 2 6 – 2y = 2 becomes y = 2 so x = 4 |
b) 2x – y = 2 becomes y = 2x – 2 2x + y = 10 becomes 2x + (2x – 2) = 10 4x = 12 becomes x = 3 so y = 4 |
c) 10x – 3y = –7
4x + 5y = 22
This is a case where we should use elimination:
Multiply the 1st equation by 5: 50x – 15y = –35
Multiply the 2nd equation by 3: 12x + 15y = 66
Add the two equations: 62x = 31 becomes x = 0.5 and y = 4.
Check the solution: 10(0.5) – 3(4) = –7 ü and 4(0.5) + 5(4) = 22 ü
d) An art supplies store sells 3 differenct kits consisting of brushes and tubes of paint. Regardless of the kit, the unit price of one brush and one tube of paint remains the same.
The "beginner's" kit with 3 brushes and 5 tubes of paint costs $20.75;
The "pro's" kit with 5 brushes and 10 tubes of paint costs $40.25;
The "expert's" kit includes 8 brushes and 20 tubes of paint.
How much does the "expert's" kit cost?
Let b = cost of 1 brush.
Let t = cost of 1 tube of paint.
3b + 5t = 20.75
5b + 10t = 40.25
Multiply 1st equation by –2 to get: – 6b – 10t = –41.50
Now add the 2 equations: – b = – 1.25 so b = $1.25
Substituting the value for b in 1st equation gives us: t = $3.40
The "expert's" kit will cost 8(1.25) + 20(3.40) = $78.00
| intro | linear systems | mixed systems | examples | practice | solutions |
1) Solve these systems algebraically:
| a) 4x + y = 3 2x – 3y = 12 |
b) 3x – 5y = 35 2.5x + y = 8.5 |
c) y = x 2 – 4x y = x + 6 |
d) x 2 + y 2 = 5 y = x – 3 |
2) Viv and Jaclyn sell magazine subscriptions. Viv's weekly salary is $100 plus a commission of $6.50 per subscription sold. Jaclyn earns $75 per week plus a $7.00 commission for each subscription sold.
a) How many subscriptions must they sell in order to earn the same amount?
(hint: let A = the amount each girl earns and x = the # of subscriptions sold,
set A for Viv = A for Jaclyn.)
b) How much do they each earn?
3) Find the points of intersection of the circle x 2 + y 2 = 125 and the line y = –x + 15.
4) The perimeter of a rectangle is 138 cm. If the length were 3 cm. longer, it would be twice the width. What are the dimensions of this rectangle? (Make a diagram immediately if not sooner !!)
| intro | linear systems | mixed systems | examples | practice | solutions |
1) Solve these systems algebraically:
| a) 4x + y = 3 2x – 3y = 12 soln: y = 3 – 4x 2x – 3( 3 – 4x) = 12 14x = 21 x = 1.5, y = –3 |
b) 3x – 5y = 35 2.5x + y = 8.5 soln: y = 8.5 – 2.5x 3x – 5(8.5 – 2.5x) = 35 15.5x = 77.5 x = 5, y = – 4 |
c) y = x 2 – 4x y = x + 6 soln: x + 6 = x 2 – 4x x 2 – 5x – 6 = 0 (x – 6)(x + 1) = 0 x = 6 or x = –1 y = 12 or y = 5 |
d) x 2 + y 2 = 5 y = x – 3 soln: x 2 + (x – 3) 2 = 5 x 2 + x 2 – 6x + 9 = 5 2x 2 – 6x + 4 = 0 x 2 – 3x + 2 = 0 (x – 2)(x – 1) = 0 x = 2 or x = 1 y = –1 or y = –2 |
2) Viv and Jaclyn sell magazine subscriptions. Viv's weekly salary is $100 plus a commission of $6.50 per subscription sold. Jaclyn earns $75 per week plus a $7.00 commission for each subscription sold.
a) How many subscriptions must they sell in order to earn the same amount?
Let x = the number of subscriptions sold
Let A v = Viv's total income = 6.5x + 100
Let A J = Jaclyn's total income = 7x + 75
set A v = A J
6.5x + 100 = 7x + 75
25 = 0.5x
50 = x
a) They must each sell 50 magazine subscriptions.
b) How much do they each earn? $425.00
check that 6.5(50) + 100 = 7(50) + 75 = $425.00-- by substituting x = 50 into the equation.
3) Find the points of intersection of the circle x 2 + y 2 = 125 and the line y = – x + 15.
substitute – x + 15 for y in x 2 + y 2 = 125
x 2 + (–x + 15) 2 = 125
x 2 + x 2 – 30x + 225 = 125
2x 2 – 30x + 100 = 0
x 2 – 15x + 50 = 0
(x – 5) (x – 10) = 0
x = 5 or x = 10
y = 10 or y = 5
The line intersects the circle at the points (5, 10) and (10, 5).
4) The perimeter of a rectangle is 138 cm. If the length were 3 cm. longer, it would be twice the width. What are the dimensions of this rectangle? (Make a diagram immediately if not sooner !!)
Let w = width
Let l = length
The perimeter = 138 so 2(l + w) = 138 becomes l + w = 69
length + 3 = twice the width so l + 3 = 2w becomes l = 2w – 3
so, 2 w – 3 + w = 69
3w = 72 becomes w = 24
l = 2w – 3 becomes l = 45
The rectangle is 24 cm. by 45 cm.
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