Hi, can someone help me and explain the steps for this problem please

Question

Before you begin to solve either system of equations below, think about what you could do to change each equation so the numbers would be more "friendly". Explain what you would do for each equation, then do it, and solve each system.

A.

y=.25x-3

y=-.2x+6

 

B.

y=1/4 - 3

y=-1/5x + 6

Answer 1

A)

 

y=.25x-3
y=-.2x+6

 

The decimal value in the first equation means that you must do careful substitution or elimination in the process of solving the system of equations.  This characteristic slightly complicates the process (although it still can be done), but it is smart to eliminate risks of errors (for example, like tying your shoes very well before starting a race).

 

To get rid of the decimal, multiply the first equation by 4 to get 4y=x-12.  Now, you may either change that to x=4y+12 and substitute for x in the second equation :

                 y = -2x + 6

                 y = -2(4y+12) + 6

                 y = -8y - 24 + 6

                 9y = -18

                 y = -2

         then (using either equation to solve for x):

                 y = -2x + 6

                 -2 = -2x + 6

                 -8 = -2x

                 4 = x

         Check:

             Does   -2 =  0.25(4) - 3    ?

                       -2 =  1 - 3   ?

                        -2 = -2   ?yes

             Does   -2 = -2(4) + 6    ?

                       -2 = - 8  +  6   ?

                       -2 = -2   ?yes

 

   Or else, you could use elimination:

          Select either x or y and make the coefficient the same (or opposite), then subtract (or add) equations:

                 8y =  2x - 24             [multiply by 8 to get 2x]

                  y  = -2x + 6

            ----------------------     [eliminate x by adding equations]

                 9y =        18

                   y =      2

             Now, either substitute or else use elimination again:

                  y =          2

                  y = -2x + 6

               -----------------      [eliminate y by subtracting equations]:

                 0 = -2x + 8

                2x = 8

                 x = 4

             Same check:

                 . . .

 

Now, for part B, the fractions cause the risk for failure, so get rid of them.  For substitution, multiply by the denominator.  For elimination, find the LCD (or LCM) [note: look these up if you don't know them] and multiply to get it as a coefficient for elimination:

         5y = 5x - 15  

          [NOTE: I assumed the missing x was a typo -- I make them all the time, so I recognize them quickly.]

         5y = -x + 30

 

  *** Hopefully, you can take it from here.

 

Answer 2

For me, I hate working with fractions or decimals so to make the equations more 'friendly',  after setting each equation equal to each other, I would multiply the equation by what's needed to make the coefficients and constants integers. This is just my personal opinion.

 

For A

y = 0.25x - 3

y = -0.2x + 6

 

As both are solved for y, you can use substitution to get

 

0.25x - 3 = -0.2x + 6

As there are at most two decimal places (hundreths), multiply everything by 100 to get

 

25x - 300 = -20x + 600

 

Now solve this equation, for x, substititute to solve for y, and please let me know if you need help solving this.

 

For B

y = 1/4x - 3

y = -1/5x + 6

 

Again use substitution to get

1/4x - 3 = -1/5x + 6

 

Now the fractions have denominators of 4 and 5 so to get rid of the fractions you can multiply the entire equation by the common denominator which is 4*5 = 20 to get

 

5x - 60 = -4x + 120

 

Now solve this equation for x then substitute and solve for y.

 

Please let me know if you need more help or any step explained more. 

 

 

 

 

Interesting note: A and B are the same problem written differently, you will get the same answer.