What is 3x^2-23x+40=0,5y^2-17y+1y=0 ?

The solution to 3x^2-23x+40=0,5y^2-17y+1y=0 is

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3x^2-23x+40=0,5y^2-17y+1y=0

3 x^{2} - 23 x + 40 = 0, 5 y^{2} - 17 y + 1 y = 0

x = \frac{8}{3}, x = \frac{8}{3}, x = 5, x = 5, y = \frac{16}{5} y = 0 y = \frac{16}{5} y = 0

Solution steps

[3 x^{2} - 23 x + 40 = 0 5 y^{2} - 17 y + 1 \cdot y = 0]

Solve

3 x^{2} - 23 x + 40 = 0 : x = \frac{8}{3}, x = 5

For

5 y^{2} - 17 y + 1 y = 0

, subsitute

x

with

\frac{8}{3} : y = \frac{16}{5}, y = 0

For

5 y^{2} - 17 y + 1 y = 0

, subsitute

x

with

5 : y = \frac{16}{5}, y = 0

Solve

5 y^{2} - 17 y + 1 y = 0 : y = 0, y = \frac{16}{5}

For

3 x^{2} - 23 x + 40 = 0

, subsitute

y

with

0 : x = 5, x = \frac{8}{3}

For

3 x^{2} - 23 x + 40 = 0

, subsitute

y

with

\frac{16}{5} : x = 5, x = \frac{8}{3}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

3 x^{2} - 23 x + 40 = 0, 5 y^{2} - 17 y + 1 y = 0

are

x = \frac{8}{3}, x = \frac{8}{3}, x = 5, x = 5, y = \frac{16}{5} y = 0 y = \frac{16}{5} y = 0

6 x^{2} + y = 1, 412 x^{2} - y = 4

abb = 6 c, c = 10

x + y = 1, 1 x y = 30

y - y^{2} y = x, x = 6

2 x + 3 y = 7, 3 x - 2 y = 4

What is 3x^2-23x+40=0,5y^2-17y+1y=0 ?
The solution to 3x^2-23x+40=0,5y^2-17y+1y=0 is