Solving Systems of Linear Equations
Mastering Systems of Linear Equations
Test your skills in solving systems of linear equations with our engaging quiz! This quiz features a variety of problems that will challenge your knowledge and enhance your understanding of both substitution and elimination methods.
Participate now and discover how well you can:
- Solve equations using substitution
- Utilize elimination techniques
- Identify solutions graphically
A system of linear equations is just what it sounds like - 2 or more equations which can describe lines in the coordinate plane.
If these lines intersect, the point at which they intersect is called the solution to the system!
A system of linear equations is just what it sounds like - 2 or more equations which can describe lines in the coordinate plane.
If these lines intersect, the point at which they intersect is called the solution to the system!
One method by which we can solve systems of equations is called substitution.
Here's an example:
y - 0.5x = 2
y + 2x = -3
By solving the bottom equation for y, we can get:
y = -2x -3
Then, we can re-write the top equation and replace the y with -2x - 3.
-2x - 3 - 0.5x = 2
Now, we can combine like terms...
-2.5 x = 5
x = -2.
Now that we have a value for x, we can plug that back into the original second equation...
y + 2(5) = -3
y + 10 = -3
y = -13.
And there you have it - substitution!
One method by which we can solve systems of equations is called substitution.
Here's an example:
y - 0.5x = 2
y + 2x = -3
By solving the bottom equation for y, we can get:
y = -2x -3
Then, we can re-write the top equation and replace the y with -2x - 3.
-2x - 3 - 0.5x = 2
Now, we can combine like terms...
-2.5 x = 5
x = -2.
Now that we have a value for x, we can plug that back into the original second equation...
y + 2(5) = -3
y + 10 = -3
y = -13.
And there you have it - substitution!
Now you try!
(questions come from Johns Hopkins CTY Honors Algebra II, KB Gardner instructor)
Now you try!
(questions come from Johns Hopkins CTY Honors Algebra II, KB Gardner instructor)
Solve this system of equations by the substitution method.
x+2y = 5
-x + y = -5
(0,5/2)
(6,11)
(5,0)
No solution
Find the solution of the system by using substitution.
x + 2y = -5
3x + y = -5
(0, −5)
(−1, −2)
(-2, -3/2)
No solution
Solve this system.
4x + 4y = 4
-12x - 12y = -13
Here's a hint: start by factoring out obvious things, but be careful of logical absurdities like 1 = 13/12...
(0,1)
(-3, 4)
(-8, 4)
No Solution
Another method of solving linear equations is by elimination.
In this method, one equation is added to or subtracted from the other to eliminate one of the variables.
For example...
4x + 3y = 2
8x - 3y = 10
Adding these two equations together will eliminate the y variable...
4x + 3y = 2
+ 8x - 3y = 10
12x = 12
x=1.
From here, we can proceed using substitution...
4(1) + 3y = 2
4 + 3y = 2
3y = -2
y = -2/3.
Another method of solving linear equations is by elimination.
In this method, one equation is added to or subtracted from the other to eliminate one of the variables.
For example...
4x + 3y = 2
8x - 3y = 10
Adding these two equations together will eliminate the y variable...
4x + 3y = 2
+ 8x - 3y = 10
12x = 12
x=1.
From here, we can proceed using substitution...
4(1) + 3y = 2
4 + 3y = 2
3y = -2
y = -2/3.
Find the solution of the system by using elimination.
2x-y = -10
x+y = -2
(0, 10)
(-18, 2)
(-4, 2)
No solution
Solve the system by the elimination method.
Hint: find a common denominator first!
(1, 4)
(2, 9/2)
(5, 3)
(4, 11/2)
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