What is 1y=x^2-5x+7,y=2x+1 ?

The solution to 1y=x^2-5x+7,y=2x+1 is

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1y=x^2-5x+7,y=2x+1

1 y = x^{2} - 5 x + 7, y = 2 x + 1

(x = 6, x = 1, y = 13 y = 3)

Solution steps

[1 \cdot y = x^{2} - 5 x + 7 y = 2 x + 1]

Subtract the equations

1 y = x^{2} - 5 x + 7

1 \cdot y - y = x^{2} - 5 x + 7 - (2 x + 1)

Simplify

0 = x^{2} - 7 x + 6

Solve

0 = x^{2} - 7 x + 6 : x = 6, x = 1

Plug the solutions

x = 6, x = 1

into

1 y = x^{2} - 5 x + 7

For

1 y = x^{2} - 5 x + 7

, subsitute

x

with

6 : y = 13

For

1 y = x^{2} - 5 x + 7

, subsitute

x

with

1 : y = 3

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

1 y = x^{2} - 5 x + 7, y = 2 x + 1

are

(x = 6, x = 1, y = 13 y = 3)

x^{2} + y = 37, 3 x + 4 y = 22

8 x + 9 y = - 5, - 8 x - 9 y = 5

\frac{1}{y} - \frac{1}{x} = \frac{1}{3}, x - 2 y = 2

- 2 x + y = 9, y = 2 x + 9

8 x^{2} - 10 x - 63 = 2, y^{2} + 33 y - 17 = 0