What are the solutions to the equation 14x^1-3x^2=8,x^1+5x^2=11 ?

The solutions to the equation 14x^1-3x^2=8,x^1+5x^2=11 are x= 2/3 ,x=4

Find the zeros of 14x^1-3x^2=8,x^1+5x^2=11

The zeros of 14x^1-3x^2=8,x^1+5x^2=11 are x= 2/3 ,x=4

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14 x^{1} - 3 x^{2} = 8, x^{1} + 5 x^{2} = 11

x = \frac{2}{3}, x = 4

Decimal Notation

x = 0.66666 \dots, x = 4

Solution steps

14 x^{1} - 3 x^{2} = 8

Refine

14 x - 3 x^{2} = 8

Move

8

to the left side

14 x - 3 x^{2} - 8 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 3 x^{2} + 14 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 ( - 3 ) ( - 8 )}{2 ( - 3 )}

1 4^{2} - 4 (- 3) (- 8) = 10

x_{1, 2} = \frac{- 14 \pm 10}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 14 + 10}{2 ( - 3 )}, x_{2} = \frac{- 14 - 10}{2 ( - 3 )}

x = \frac{- 14 + 10}{2 ( - 3 )} : \frac{2}{3}

x = \frac{- 14 - 10}{2 ( - 3 )} : 4

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = 4

y, y^{2} = x (x^{2} - 3)

a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0

100 t^{2} - 192 t + 256 = 0

\frac{a}{b} = k^{2}

x, j = (4 + 3 x) (8 x + s)

What are the solutions to the equation 14x^1-3x^2=8,x^1+5x^2=11 ?
The solutions to the equation 14x^1-3x^2=8,x^1+5x^2=11 are x= 2/3 ,x=4
Find the zeros of 14x^1-3x^2=8,x^1+5x^2=11
The zeros of 14x^1-3x^2=8,x^1+5x^2=11 are x= 2/3 ,x=4