Algebraically solve the system of equations shown below. Note that you can use either factoring or the quadratic formula to find the X – coordinates, but the quadratic formula is probably easier.

6 x 2 + 19 x − 15 = − 12 x + 15 6 x 2 + 31 x − 30 = 0 6 x 2 + 36 x − 5 x − 30 = 0 6 x ( x + 6 ) − 5 ( x + 6 ) = 0 ( 6 x − 5 ) ( x + 6 ) = 0 x = 6 5 ​ ∨ x = − 6 y = − 12 ⋅ 6 5 ​ + 15 ∨ y = − 12 ⋅ ( − 6 ) + 15 y = − 10 + 15 ∨ y = 72 + 15 y = 5 ∨ y = 87 x = 6 5 ​ ∧ y = 5 x = − 6 ∧ y = 87

Answered by konrad509 | 2024-06-10

y = 6 x 2 + 19 x − 15 y = − 12 x + 15 6 x 2 + 19 x − 15 = − 12 x + 15 6 x 2 + 19 x + 12 x − 15 − 15 = 0 6 x 2 + 31 x − 30 = 0 a = 6 b = 31 c = − 30 x = 2 a − b ± b 2 − 4 a c ​ ​ = 2 ⋅ 6 − 31 ± 3 1 2 − 4 ⋅ 6 ⋅ ( − 30 ) ​ ​ = 12 − 31 ± 1681 ​ ​ = 12 − 31 ± 41 ​ x = 12 − 31 − 41 ​ or x = 12 − 31 + 41 ​ x = − 6 or x = 6 5 ​
y = − 12 ⋅ ( − 6 ) + 15 or y = − 12 ⋅ 6 5 ​ + 15 y = 87 or y = 5 the answer: x = − 6 , y = 87 ​ or x = 6 5 ​ , y = 5 ​

Answered by naǫ | 2024-06-10

The system of equations is solved using the quadratic formula, resulting in two solutions: ( 6 5 ​ , 5 ) and (-6, 87). First, we rearranged the equation and found the discriminant. Then we calculated the x-values and substituted them to find corresponding y-values.
;

Answered by konrad509 | 2024-12-18