Learn how to use the Algebra Calculator to solve systems of equations.
Example Problem
Solve the following system of equations:
x+y=7, x+2y=11
How to Solve the System of Equations in Algebra Calculator
First go to the Algebra Calculator main page.
Type the following:
- The first equation x+y=7
- Then a comma ,
- Then the second equation x+2y=11
Try it now: x+y=7, x+2y=11
Clickable Demo
Try entering x+y=7, x+2y=11 into the text box.
After you enter the system of equations, Algebra Calculator will solve the system x+y=7, x+2y=11 to get x=3 and y=4.
More Examples
Here are more examples of how to solve systems of equations in Algebra Calculator. Feel free to try them now.
- Solve y=x+3, y=2x+1: y=x+3, y=2x+1
- Solve 2x+3y=5, x+y=4: 2x+3y=5, x+y=4
Need Help?
Please feel free to Ask MathPapa if you run into problems.
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Systems of Linear equations:
A system of linear equations is just a set of two or more linear equations.
In two variables ( x and y ) , the graph of a system of two equations is a pair of lines in the plane.
There are three possibilities:
- The lines intersect at zero points. (The lines are parallel.)
- The lines intersect at exactly one point. (Most cases.)
- The lines intersect at infinitely many points. (The two equations represent the same line.)
How to Solve a System Using The Substitution Method
- Step 1 : First, solve one linear equation for y in terms of x .
- Step 2 : Then substitute that expression for y in the other linear equation. You'll get an equation in x .
- Step 3 : Solve this, and you have the x -coordinate of the intersection.
- Step 4 : Then plug in x to either equation to find the corresponding y -coordinate.
Note 1 : If it's easier, you can start by solving an equation for x in terms of y , also – same difference!
Example:
Solve the system { 3 x + 2 y = 16 7 x + y = 19
Solve the second equation for y .
y = 19 − 7 x
Substitute 19 − 7 x for y in the first equation and solve for x .
3 x + 2 ( 19 − 7 x ) = 16 3 x + 38 − 14 x = 16 − 11 x = − 22 x = 2
Substitute 2 for x in y = 19 − 7 x and solve for y .
y = 19 − 7 ( 2 ) y = 5
The solution is ( 2 , 5 ) .Note 2 : If the lines are parallel, your x -terms will cancel in step 2 , and you will get an impossible equation, something like 0 = 3 .
Note 3 : If the two equations represent the same line, everything will cancel in step 2 , and you will get a redundant equation, 0 = 0 .
The substitution method is a way to solve the system of equations which requires calculating one of the variables from one equation and substituting it into the other equation.
The initial system is:
#{(2x+y=13),(5x-2y=10):}# ##
You can easily calculate #y# from the first equation:
#y=-2x+13#
Now if you substitute this value in the second equation you get the equation with one variable only (#x#):
#5x-2*(-2x+13)=10#
#5x+4x-26=10#
#9x=36 =>x=4#
Now you can substitute the calculated value of #x# into any of the equations:
#y=-2*4+13=>y=-8+13 =>y=5#
Finally we can write the answer:
This system has one solution:
#{(x=4),(y=5):}#
#2x+y=13to(1)#
#5x-2y=10to(2)#
#"rearrange equation "(1)" to give y in terms of x"#
#rArry=13-2xto(3)#
#color(blue)"substitute "y=13-2x" into equation "(2)#
#rArr5x-2(13-2x)=10#
#rArr5x-26+4x=10#
#rArr9x-26=10#
#"add "26" to both sides"#
#9xcancel(-26)cancel(+26)=10+26#
#rArr9x=36#
#"divide both sides by "9#
#(cancel(9) x)/cancel(9)=36/9#
#rArrx=4#
#"substitute this value into equation "(3)" and evaluate for y"#
#rArry=13-(2xx4)=13-8=5#
#color(blue)"As a check"#
#"substitute these values into equation "(2)#
#(5xx4)-(2xx5)=20-10=10larrcolor(blue)"True"#
#rArr"point of intersection "=(4,5)#
graph{(y+2x-13)(y-5/2x+5)=0 [-10, 10, -5, 5]}
#2x+y=13 ; (1) , 5x-2y=10; (2)#. From equation (1)
we get #y=13-2x #. Substituting #y=13-2x # in
equation (2) we get #5x - 2(13-2x)=10 # or
#5x - 26+4x =10 or 9x = 10+26 or 9x=36 or x=4#.
Substituting #x=4# in equation (1) we get ,
#2*4+y=13 :. y = 13-8 or y=5 :. x=4 , y=5#
Solution: #x=4 , y=5# [Ans]