Question

Answer for 1A. With explanation

Answer 1

Answer: x = 3, y = -1   →   (3, -1)

Step-by-step explanation:

When solving a system, you want the coefficients of one of the variables to be opposites (same number but different signs).

Multiply each equation by a number that will create one of the coefficients to be opposites.  Then add the two equations to eliminate one of the variables.

2x  - 5y = 11      →    3(2x - 5y = 11 )      →    6x - 15y = 33

-3x + 4y = -13    →    2(-3x + 4y = -13)   →   -6x + 8y = -26

                                                                          -7y =   7

                                                                        ÷-7    ÷-7  

                                                                             y = -1

Now input y = -1 into either of the equations to solve for x

2x -  5y  = 11

2x - 5(-1) = 11

2x  + 5   = 11

2x          =  6

 x          =   3

So, the solution to the system is: x = 3, y = -1   →   (3, -1)

Using the table method, input values for x into each of the equations to solve for y.  If both equations result in the same y-value, then that is your solution.

Try x = 0:

 1st equation                                2nd equation  

  2x - 5y₁ = 11                                  -3x + 4y₂ = -13

2(0) - 5y₁ = 11                                 -3(0) + 4y₂ = -13

   0  - 5y₁ = 11                                     0   + 4y₂ = -13

           y₁ = $-\dfrac{11}{5}$                                       y₂ = $-\dfrac{13}{4}$

                         y₁ ≠ y₂ so this is not a solution

Do the same for x = 1, x = 2, and x = 3.  The result is the table below.

$\begin{array}{c|c|c|c|c}x&y_1&y_2&(x,y_1)&(x,y_2)\\0&-\dfrac{11}{5}&-\dfrac{13}{4}&(0,-\dfrac{11}{5})&(0,-\dfrac{13}{4})\\\\1&-\dfrac{9}{5}&-\dfrac{5}{2}&(0,-\dfrac{9}{5})&(0,-\dfrac{5}{2})\\\\2&-\dfrac{7}{5}&-\dfrac{7}{4}&(0,-\dfrac{7}{5})&(0,-\dfrac{7}{4})\\\\3&-1&-1&(3,-1_&(3,-1)\end{array}\right]$