-x + y = 2
4x - 3y = -7
-x + y = 2 (add x to both sides)
y = x + 2 (substitute x + 2 for y into the second equation of the system)
4x - 3y = -7
4x - 3(x + 2) = -7 (solve for x)
4x - 3x - 6 = -7
x - 6 = -7 (add 6 to both sides)
x = -1
y = x + 2 (substitute -1 for x)
y = -1 + 2
y = 1
Thus (-1, 1) is the solution of the given system of the equations.
x+y=5
isolate
by elimination,substitution or through the matrix method.
2x + 2y = 44x + y = 1There are many methods you can use to solve this system of equations (graphing, elimination, substitution, matrices)...but no matter what method you use, you should get x = -1/3 and y = 7/3.
Substitution is a way to solve without graphing, and sometimes there are equations that are impossible or very difficult to graph that are easier to just substitute. Mostly though, it is a way to solve if you have no calculator or cannot use one (for a test or worksheet).
Use the substitution method to solve the system of equations. Enter your answer as an ordered pair.y = 2x + 5 x = 1
(2,3)
x+y=5
isolate
2x + 2y = 44x + y = 1There are many methods you can use to solve this system of equations (graphing, elimination, substitution, matrices)...but no matter what method you use, you should get x = -1/3 and y = 7/3.
by elimination,substitution or through the matrix method.
You can solve lineaar quadratic systems by either the elimination or the substitution methods. You can also solve them using the comparison method. Which method works best depends on which method the person solving them is comfortable with.
Substitution is a way to solve without graphing, and sometimes there are equations that are impossible or very difficult to graph that are easier to just substitute. Mostly though, it is a way to solve if you have no calculator or cannot use one (for a test or worksheet).
2x+7y=29 x=37-8y
You'd need another equation to sub in
If: x+y = 4 and y = 2x+1 Then: 4-x = 2x+1 => 3 = 3x => 1 = x So by substitution: x = 1 and y = 3
That's exactly the purpose of the substitution method ... to get an equation with one less variable. When you have it, you solve it for the variable that's left.