7-1 Solving Systems by Graphing
Two or more linear equations together make up a system of linear equations. The solution to a system is the ordered pair(s) that work, or are true in both equations. Visually, this is the point where they cross.
If you want to find the system solution visually, you need to graph both equations on the same Cartesian plane (same graph). This method works well when the equations for the graphs are given in slope intercept form, and the solutions are at an evenly distinguishable point.
If you graph two lines and they never cross (they are parallel) then there will never be a point that they have in common and we say there is no solution.
If you graph two lines and they lie on top of each other, we say that the two lines coincide, and they have all solutions in common, or infinitely many solutions in common (the identity).
Example:
7-2 Solving Systems Using Substitution
The substitution method works best when one of the variables is already solved for in one of the equations. In other words, if you have a y= or an x = as one of your equations. Substituting this value into the other equation is fairly simple.
I suggest re-writing the equation you are substituting into using parenthesis.
So if you were given:
Y = 3x + 2 and -2x + 5y = 10 you would substitute the first into the second so re-write the second with parenthesis around the y, since that is what you will be replacing.
Y = 3x + 2 and -2x + 5(y) = 36
Then put the value of y, (3x + 2), in place of the y in the second equation and solve for x.
-2x + 5(3x + 2) = 36
-2x + 15x + 10 = 36
13x + 10 = 36
13x = 26
X = 2
Then substitute that into the first equation to solve for y
Y = 3(x) + 2
Y = 3(2) + 2
Y = 6 + 2
Y = 8
The ordered pair solution then is (2, 8)
7-3 Solving Systems Using Elimination
The elimination method is best used when the substitution method is difficult, or when both equations are in the exact same format.
The idea is to get leading coefficients on one of the variables to be opposite so that when you combine the two equations, one of them will be eliminated. This is also often called the add/subtract method.
Example:
3x – 7y = -32
1x + 7y = 48 Notice that the coefficients on the “y” variable are opposite
When we add the equations together we get:
3x – 7y = -32
1x + 7y = 48
4x = 16
x = 4
Substitute this value into either equation to solve for y
3(4) – 7y = -32
12 – 7y = -32
-7y = -44
Y = 6 2/7
So the ordered pair is (4, 6 2/7)
You might have to multiply one or both of the equations by something in order to get the leading coefficients the same with opposite signs.
Example:
4x + 5y = 15
6x – 4y = 11
If I multiply the top by 3 & the bottom by -2, the leading coefficients on the x will be the same with opposite signs.
3(4x + 5y = 15) 12x + 15y = 45
-2(6x – 4y = 11) -12x + 8y = -22
23y = 23
Y = 1
Substitute the value of y in and solve for x.
4x + 5(1) = 15 OR 6x – 4(1) = 11
4x + 5 = 15 6x – 4 = 11
4x = 10 6x = 15
x = 2 ½ x = 2 ½
Either way you get the same x value, so the ordered pair is (2 ½ , 1)
7-4 Applications of Linear Systems
The biggest hang up for most people is determining which method to use when solving systems. Honestly, any method will work with any problem, although different methods are easier in different situations. If you are only comfortable with one, stick with that one – just be cautious that the graphing method is very difficult to get exact answers when the points are not gridded points (easily identifiable). I would pick one of the others to become a pro at.
Graphing method: Used best when given equations in slope intercept form
Substitution Method: Used best when one variable is already solved for or can easily be solved for (if it has a leading coefficient of one)
Elimination Method: Used best when two equations are in the same format.
If none of the above applies, you will need to do some rearranging with your equations so that one of the above will be applicable.
Example:
Suppose I fly from Rapid City to Minneapolis and it takes 1 hour 45 minutes to go 750 miles, but I’m flying into the wind. At the same time, Ms Grant is flying from Minneapolis to Rapid on a plane traveling at the same speed, but it only takes 1 hour 20 minutes because she is flying with the wind. Find the speed of the plane and the speed of the wind.
I know that rate times time is distance, so I will write an equation for each of us using that:
Let R = Plane Speed (rate)
Let W = Wind Speed
Me:
(R-W) * 1.75 = 750 I used 1.75 because 45 min is ¾ of an hour, or .75
I subtracted the wind speed because flying INTO
the wind (head wind) will slow me down
Ms Grant:
(R + W) * 1.333 = 750 I used 1.333 because 20 min is 1/3 of an hour or .333
I added the wind speed because traveling with the
wind (tail wind) speeds you up
I set it equal to 750 because the distance is the same
between both cities no matter which way you travel.
Distribute and line up each equation:
1.75R – 1.75 W = 750
1.333R + 1.333W = 750
I recognize this might be easier with improper fractions:
7/4 R – 7/4 W = 750
4/3R + 4/3W = 750
I multiply the top equation by 4 and the bottom by 3 to get rid of the fractions:
7R – 7W = 3000
4R + 4W = 2250
Now I will multiply the top equation by 4 and the bottom by 7 to get opposite leading coefficients on the W and eliminate:
28R – 28W = 12000
28R + 28W = 15750
56R = 27750
R = 495.5 miles per hour
I now know the plane speed and can use it to figure the wind speed
7/4 (495.5) – 7/4 W = 750
13885/16 – 28/16 W = 750 Notice I changed the fractions all to 16th’s
13885 – 28 W = 12000 Multiplied everything by 16 to get rid of fractions
-28W = -1885
W = 67.3
Now I know the wind speed as well, it is 67.3 miles per hour.
7-5 Linear Inequalities
Graphing Linear Inequalities, for all practical purposes, is the same as graphing linear equations. The solution to a linear inequality is the graph, we often graph on a coordinate plane and indicate greater than by shading above the line and less than by shading below the line. A solid line indicates equality, a dashed line indicates only greater than or less than, but NOT equal to.
Since we are graphing, it is often easiest to get the inequality into slope intercept form, (y = mx + b) this keeps the sign accurate. If we have a y < 3x + 2, we know we are shading below a solid line because we read it from the standpoint of the positive y.
Example:
Graph y < 3x + 2
Graph 2x + 3y > 6
This equation is in standard form which means we can easily get the intercepts.
The x intercept is at 3 and the y intercept is at 2. Plot these points, connect the dots with a dashed line, and shade above the lines since we have a positive y > . If it had been a negative y > we would have shaded below. (Because multiplying by the negative to get a positive y would change the sign.)
7-6 Systems of Linear Inequalities
Systems of Inequalities are done the same as systems of linear equations, with the exception of the dashed vs. solid lines and the shading above and below. The solution to a system of linear inequalities is the graph where both equations have been graphed and shaded, and they overlap. Any point in this overlapping region should be a solution to both linear inequalities.
Example:
Solve the system of Linear Inequalities
Y > ½ x + 1 This is in slope intercept form and ready to graph
4x + 2y < 8 This is standard form, get intercepts – x at 2 & y at 4
Two or more linear equations together make up a system of linear equations. The solution to a system is the ordered pair(s) that work, or are true in both equations. Visually, this is the point where they cross.
If you want to find the system solution visually, you need to graph both equations on the same Cartesian plane (same graph). This method works well when the equations for the graphs are given in slope intercept form, and the solutions are at an evenly distinguishable point.
If you graph two lines and they never cross (they are parallel) then there will never be a point that they have in common and we say there is no solution.
If you graph two lines and they lie on top of each other, we say that the two lines coincide, and they have all solutions in common, or infinitely many solutions in common (the identity).
Example:
7-2 Solving Systems Using Substitution
The substitution method works best when one of the variables is already solved for in one of the equations. In other words, if you have a y= or an x = as one of your equations. Substituting this value into the other equation is fairly simple.
I suggest re-writing the equation you are substituting into using parenthesis.
So if you were given:
Y = 3x + 2 and -2x + 5y = 10 you would substitute the first into the second so re-write the second with parenthesis around the y, since that is what you will be replacing.
Y = 3x + 2 and -2x + 5(y) = 36
Then put the value of y, (3x + 2), in place of the y in the second equation and solve for x.
-2x + 5(3x + 2) = 36
-2x + 15x + 10 = 36
13x + 10 = 36
13x = 26
X = 2
Then substitute that into the first equation to solve for y
Y = 3(x) + 2
Y = 3(2) + 2
Y = 6 + 2
Y = 8
The ordered pair solution then is (2, 8)
7-3 Solving Systems Using Elimination
The elimination method is best used when the substitution method is difficult, or when both equations are in the exact same format.
The idea is to get leading coefficients on one of the variables to be opposite so that when you combine the two equations, one of them will be eliminated. This is also often called the add/subtract method.
Example:
3x – 7y = -32
1x + 7y = 48 Notice that the coefficients on the “y” variable are opposite
When we add the equations together we get:
3x – 7y = -32
1x + 7y = 48
4x = 16
x = 4
Substitute this value into either equation to solve for y
3(4) – 7y = -32
12 – 7y = -32
-7y = -44
Y = 6 2/7
So the ordered pair is (4, 6 2/7)
You might have to multiply one or both of the equations by something in order to get the leading coefficients the same with opposite signs.
Example:
4x + 5y = 15
6x – 4y = 11
If I multiply the top by 3 & the bottom by -2, the leading coefficients on the x will be the same with opposite signs.
3(4x + 5y = 15) 12x + 15y = 45
-2(6x – 4y = 11) -12x + 8y = -22
23y = 23
Y = 1
Substitute the value of y in and solve for x.
4x + 5(1) = 15 OR 6x – 4(1) = 11
4x + 5 = 15 6x – 4 = 11
4x = 10 6x = 15
x = 2 ½ x = 2 ½
Either way you get the same x value, so the ordered pair is (2 ½ , 1)
7-4 Applications of Linear Systems
The biggest hang up for most people is determining which method to use when solving systems. Honestly, any method will work with any problem, although different methods are easier in different situations. If you are only comfortable with one, stick with that one – just be cautious that the graphing method is very difficult to get exact answers when the points are not gridded points (easily identifiable). I would pick one of the others to become a pro at.
Graphing method: Used best when given equations in slope intercept form
Substitution Method: Used best when one variable is already solved for or can easily be solved for (if it has a leading coefficient of one)
Elimination Method: Used best when two equations are in the same format.
If none of the above applies, you will need to do some rearranging with your equations so that one of the above will be applicable.
Example:
Suppose I fly from Rapid City to Minneapolis and it takes 1 hour 45 minutes to go 750 miles, but I’m flying into the wind. At the same time, Ms Grant is flying from Minneapolis to Rapid on a plane traveling at the same speed, but it only takes 1 hour 20 minutes because she is flying with the wind. Find the speed of the plane and the speed of the wind.
I know that rate times time is distance, so I will write an equation for each of us using that:
Let R = Plane Speed (rate)
Let W = Wind Speed
Me:
(R-W) * 1.75 = 750 I used 1.75 because 45 min is ¾ of an hour, or .75
I subtracted the wind speed because flying INTO
the wind (head wind) will slow me down
Ms Grant:
(R + W) * 1.333 = 750 I used 1.333 because 20 min is 1/3 of an hour or .333
I added the wind speed because traveling with the
wind (tail wind) speeds you up
I set it equal to 750 because the distance is the same
between both cities no matter which way you travel.
Distribute and line up each equation:
1.75R – 1.75 W = 750
1.333R + 1.333W = 750
I recognize this might be easier with improper fractions:
7/4 R – 7/4 W = 750
4/3R + 4/3W = 750
I multiply the top equation by 4 and the bottom by 3 to get rid of the fractions:
7R – 7W = 3000
4R + 4W = 2250
Now I will multiply the top equation by 4 and the bottom by 7 to get opposite leading coefficients on the W and eliminate:
28R – 28W = 12000
28R + 28W = 15750
56R = 27750
R = 495.5 miles per hour
I now know the plane speed and can use it to figure the wind speed
7/4 (495.5) – 7/4 W = 750
13885/16 – 28/16 W = 750 Notice I changed the fractions all to 16th’s
13885 – 28 W = 12000 Multiplied everything by 16 to get rid of fractions
-28W = -1885
W = 67.3
Now I know the wind speed as well, it is 67.3 miles per hour.
7-5 Linear Inequalities
Graphing Linear Inequalities, for all practical purposes, is the same as graphing linear equations. The solution to a linear inequality is the graph, we often graph on a coordinate plane and indicate greater than by shading above the line and less than by shading below the line. A solid line indicates equality, a dashed line indicates only greater than or less than, but NOT equal to.
Since we are graphing, it is often easiest to get the inequality into slope intercept form, (y = mx + b) this keeps the sign accurate. If we have a y < 3x + 2, we know we are shading below a solid line because we read it from the standpoint of the positive y.
Example:
Graph y < 3x + 2
Graph 2x + 3y > 6
This equation is in standard form which means we can easily get the intercepts.
The x intercept is at 3 and the y intercept is at 2. Plot these points, connect the dots with a dashed line, and shade above the lines since we have a positive y > . If it had been a negative y > we would have shaded below. (Because multiplying by the negative to get a positive y would change the sign.)
7-6 Systems of Linear Inequalities
Systems of Inequalities are done the same as systems of linear equations, with the exception of the dashed vs. solid lines and the shading above and below. The solution to a system of linear inequalities is the graph where both equations have been graphed and shaded, and they overlap. Any point in this overlapping region should be a solution to both linear inequalities.
Example:
Solve the system of Linear Inequalities
Y > ½ x + 1 This is in slope intercept form and ready to graph
4x + 2y < 8 This is standard form, get intercepts – x at 2 & y at 4