Answer:
The solutions I found to this system of equations are (1, 0) and (⁸/₁₅, -7).
I keep second guessing and recalculating the second point there, so I'd suggest double-checking it by going through the work yourself. The first one checks out though.
Step-by-step explanation:
Let's solve this by substitution, where one equation is solved for one variable and substituted in to the other. As the latter equation is already solved for y, we can simply substitute it into the other:
given:
y = 3x² - 4x + 1
3x - 5y = 3
then:
3x - 5(3x² - 4x + 1) = 3
Which we can then simplify and solve:
3x - 15x² + 20x - 5 = 3
-3x + 15x² - 20x + 5 = -3
15x² - 23x = -8
Now we can find find the solution by completing the square:
15(15x² - 23x) = 15(-8)
225x² - 345x = -120
225x² - 345x + (23/2)² = -120 + (23/2)²
225x² - 345x + 529/4 = -120 + 529/4
(15x - 11.5)² = -120 + 132.25
(15x - 11.5)² = 12.25
15x - 11.5 = ± √12.25
15x - 11.5 = ± 3.5
15x = 11.5 ± 3.5
x = (11.5 ± 3.5) / 15
x = 15 / 15, and 8 / 15
x = 1, and 0.5333...
Now that we have x, we can solve for y.
given: 3x - 5y = 3
for x = 1:
3 - 5y = 3
y = 0
for x = ⁸/₁₅:
3(⁸/₁₅) - 5y = 3
5y = 3(⁸/₁₅) - 3
5y = ²⁴/₁₅ - ⁴⁵/₁₅
5y = -²¹/₁₅
y = -²¹/₃
y = -7
So the solutions to this system of equations are (1, 0) and (⁸/₁₅, -7).
