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 = [tex]-\dfrac{11}{5}[/tex] y₂ = [tex]-\dfrac{13}{4}[/tex]
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.
[tex]\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][/tex]
