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what type of system is x+y=11 and what is the answer

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    x + y = 11   ==>   This is a linear equation in standard form, it is not a system. A system of equations consists of two or more equations.

You can rearrange the linear equation to put it in slope-intercept form, y = mx + b, where m is the slope of the line and b is the y-intercept (the y-intercept is where the line intercepts with the y-axis; in other words, it is what y is when x=0). To do so, subtract x from both sides of the equation:

          x + y = 11

        - x          - x

     _______________

         0 + y = 11 - x      

               y = -x + 11     ==>     slope:  m = -1     ,      y-intercept:  b = 11

Thus, one solution to this equation is (0, 11). That is, y = 11 when x = 0.

For the next problem:        3x + 3y = 6     and      2x + y = 2

   This is a system of linear equations (with two variables, x and y). To solve for this system of linear equations, you need to find where these two equations intersect. One way of solving for the system is by the method of elimination, where you manipulate one or both equations in a way that will eliminate one of the variables when the equations are added to one another so as to solve for the other variable. Once you have solved for one of these variables, you can solve for the other variable.

First, notice that the first equation can be simplified as follows:

        3x + 3y = 6                

         factor out a 3 from the left hand side of the equation

        3*(x + y) = 6              

         divide both sides of the equation by 3      

       (3*(x + y))/3 = 6/3

        x + y = 2 

So, the system of linear equations consists of the following equations:

        x + y = 2

      2x + y = 2

Now we can solve for the system using the method of elimination. Let's first eliminate the y variable by multiplying the first equation by -1:

   (-1)*(x + y = 2)   ==>  (-1)*(x) + (-1)*(y) = (-1)*(2)  ==>   -x - y = -2 

Now we add this equation to the second equation above:

         -x - y = -2

  +    2x + y = 2

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       -x + 2x - y + y = -2 + 2    ==>    1x + 0y = 0    ==>    x = 0  

Since we have solved for x, we plug it back into one of the original equations to solve for y. Let's plug it back into the second equation:

        2x + y = 2   ,      given that   x = 0

        2(0) + y = 2     ==>     0 + y = 2     ==>     y = 2    

Thus, the solution to this system of linear equations is:   x = 0 , y = 2    OR   (0, 2)

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