STANDARD 9:
SOLVING SYSTEMS OF EQUATIONS
Systems of equations- A set of 2 or more equations with 2 variables.
Solution to a system of equations- The solution is a point (x, y) where the lines intersect.
There are 3 methods to solve a system of equations:
- Graphing
- Substitution
- Elimination (sometimes called addition or subtraction)
Watch this video describing the 3 methods: http://www.virtualnerd.com/algebra-2/linear-systems/equations-solution-methods.php
Solution to a system of equations- The solution is a point (x, y) where the lines intersect.
There are 3 methods to solve a system of equations:
- Graphing
- Substitution
- Elimination (sometimes called addition or subtraction)
Watch this video describing the 3 methods: http://www.virtualnerd.com/algebra-2/linear-systems/equations-solution-methods.php
Graphing to Solve
How to solve a system of equations by Graphing
1.Put the equations in slope intercept form (y=mx+b)
2. Graph the lines (start with the y-intercept and move with the slope)- see st 6/7 for help!
3. Find the point where the lines intersect/cross (x,y). This is your solution
See one example here:http://www.virtualnerd.com/algebra-2/linear-systems/graphing/solve-by-graphing/equations-solution-by-graphing
1.Put the equations in slope intercept form (y=mx+b)
2. Graph the lines (start with the y-intercept and move with the slope)- see st 6/7 for help!
3. Find the point where the lines intersect/cross (x,y). This is your solution
See one example here:http://www.virtualnerd.com/algebra-2/linear-systems/graphing/solve-by-graphing/equations-solution-by-graphing
Solving Using Substitution
How to solve a system by Substitution.
1. Solve one equation for one variable
2. Put parenthesis around the expression that the variable equals and substitute it into the other equation
3. Simplify and solve for the variable
4. Substitute this value into one of the original equations and solve for the other variable to get your solution (x,y)
5. Check your work by substituting the x and y into the other original equation
Example: Solve the following system by substitution.
2x – 3y = –2
4x + y = 24
The idea here is to solve one of the equations for one of the variables, and plug this into the other equation. It does not matter which equation or which variable you pick. There is no right or wrong choice; the answer will be the same, regardless. But — some choices may be better than others. In this case, can you see that it would probably be simplest to solve the second equation for "y =", since there is already a y floating around loose in the middle there? I could solve the first equation for either variable, but I'd get fractions, and solving the second equation for x would also give me fractions. It wouldn't be "wrong" to make a different choice, but it would probably be more difficult. Being lazy, I'll solve the second equation for y:
4x + y = 24
-4x -4x
y = –4x + 24
Now-plug this in ("substitute it") for "y" in the first equation, and solve for x:
2x – 3(–4x + 24) = –2
2x + 12x – 72 = –2
14x = 70
x = 5
Now I can plug this x-value back into either equation, and solve for y. But since I already have an expression for "y =", it will be simplest to just plug into this:
y = –4(5) + 24
= –20 + 24
= 4
Then the solution is (x, y) = (5, 4).
Watch this video for an extra example: http://www.virtualnerd.com/algebra-2/linear-systems/solve-algebraically/solve-by-substitution/equations-solution-by-substitution
1. Solve one equation for one variable
2. Put parenthesis around the expression that the variable equals and substitute it into the other equation
3. Simplify and solve for the variable
4. Substitute this value into one of the original equations and solve for the other variable to get your solution (x,y)
5. Check your work by substituting the x and y into the other original equation
Example: Solve the following system by substitution.
2x – 3y = –2
4x + y = 24
The idea here is to solve one of the equations for one of the variables, and plug this into the other equation. It does not matter which equation or which variable you pick. There is no right or wrong choice; the answer will be the same, regardless. But — some choices may be better than others. In this case, can you see that it would probably be simplest to solve the second equation for "y =", since there is already a y floating around loose in the middle there? I could solve the first equation for either variable, but I'd get fractions, and solving the second equation for x would also give me fractions. It wouldn't be "wrong" to make a different choice, but it would probably be more difficult. Being lazy, I'll solve the second equation for y:
4x + y = 24
-4x -4x
y = –4x + 24
Now-plug this in ("substitute it") for "y" in the first equation, and solve for x:
2x – 3(–4x + 24) = –2
2x + 12x – 72 = –2
14x = 70
x = 5
Now I can plug this x-value back into either equation, and solve for y. But since I already have an expression for "y =", it will be simplest to just plug into this:
y = –4(5) + 24
= –20 + 24
= 4
Then the solution is (x, y) = (5, 4).
Watch this video for an extra example: http://www.virtualnerd.com/algebra-2/linear-systems/solve-algebraically/solve-by-substitution/equations-solution-by-substitution
The Elimination Method
How to solve a system by Elimination:
1. Line the equations up in standard form
2. Make sure one variable has opposite coefficients (if not, multiply one whole equation by a number to make opposite coefficients).
3. Add straight down
4. Solve for the remaining variable
5. Substitute this value into one of the original equations to solve for the other variable and getting the solution (x, y)
6. Check your work by substituting the x and y into the other original equation
Example: Solve the following system using elimination
2x + y = 9
3x – y = 16
Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:
2x + y = 9
3x – y = 16
5x = 25
Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y.
The first equation has smaller numbers, so I'll back-solve in that one:
2(5) + y = 9
10 + y = 9
y = –1
Then the solution is (x, y) = (5, –1).
Watch a more challenging example:http://www.virtualnerd.com/algebra-1/systems-equations-inequalities/equations-solution-by-elimination-multiplication.php
1. Line the equations up in standard form
2. Make sure one variable has opposite coefficients (if not, multiply one whole equation by a number to make opposite coefficients).
3. Add straight down
4. Solve for the remaining variable
5. Substitute this value into one of the original equations to solve for the other variable and getting the solution (x, y)
6. Check your work by substituting the x and y into the other original equation
Example: Solve the following system using elimination
2x + y = 9
3x – y = 16
Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:
2x + y = 9
3x – y = 16
5x = 25
Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y.
The first equation has smaller numbers, so I'll back-solve in that one:
2(5) + y = 9
10 + y = 9
y = –1
Then the solution is (x, y) = (5, –1).
Watch a more challenging example:http://www.virtualnerd.com/algebra-1/systems-equations-inequalities/equations-solution-by-elimination-multiplication.php
System of Inequalities
System of inequalities: Two or more inequalities that share a solution
Solution to a system of inequalities- The solution is all of the points (usually a shaded area) that make both of the inequalities true.
** Remember if you multiply or divide by a negative number, you need to flip the inequality sign over
To solve a system of inequalities:
1. Put both equations in slope intercept form (y = mx +b)
2. Graph one line at a time (make sure you use the correct line, either dotted if the sign is < or > and solid if < or >)
3. Shade above the line if it is greater than, shade below the line if it is less than
4. Graph and shade the other line
5. Where the shaded areas over lap is the solution (shade this part darkest)
Watch this video example: http://www.virtualnerd.com/pre-algebra/linear-functions-graphing/inequality-solve-by-graphing.php
Solution to a system of inequalities- The solution is all of the points (usually a shaded area) that make both of the inequalities true.
** Remember if you multiply or divide by a negative number, you need to flip the inequality sign over
To solve a system of inequalities:
1. Put both equations in slope intercept form (y = mx +b)
2. Graph one line at a time (make sure you use the correct line, either dotted if the sign is < or > and solid if < or >)
3. Shade above the line if it is greater than, shade below the line if it is less than
4. Graph and shade the other line
5. Where the shaded areas over lap is the solution (shade this part darkest)
Watch this video example: http://www.virtualnerd.com/pre-algebra/linear-functions-graphing/inequality-solve-by-graphing.php
Word Problems!
There will DEFINITELY be systems of equations word problems on the CST and on your final... be ready!
How to solve a systems word problem:
1. Identify the variables
2. Write the equations
3. Solve the system
4. Check your work!
Watch this video example: http://www.virtualnerd.com/algebra-2/linear-systems/two-simultaneous-equations-word-problem.php
How to solve a systems word problem:
1. Identify the variables
2. Write the equations
3. Solve the system
4. Check your work!
Watch this video example: http://www.virtualnerd.com/algebra-2/linear-systems/two-simultaneous-equations-word-problem.php
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