Question

using the properties of proportion,solve the following equation for x : x2 + 2x/2x+4 = y2 + 3y/3y + 9 ?​

Answer 1

Answer:

+ 2y = 5

Put the value of x = 0 in the equation x + 2y = 5

0 + 2y = 5

2y = 5

2y/2 = 5/2

y = 5/2

(0, 5/2)

Put the value of y = 0,

x + 2 × 0 = 5

x = 5

(5, 0)

Put the value of x = 2,

2 + 2y = 5

2 – 2 + 2y = 5 - 2

2y = 3

2y/2 = 3/2

y = 3/2

(2, 3/2)

Put the value of y = 2,

x + 2 × 2 = 5

x + 4 = 5

x + 4 – 4 = 5 – 4

x = 1

(1, 2)

Answer 2

Answer:

x = $\sqrt{y^{2}+5 }$

Step-by-step explanation:

We know 2x/2x = 1 and 3y/3y = 1, so we can write

$x^2 + 5 = y^2 + 10$

We subtract 5 from both sides and get

$x^2 = y^2 + 5$

Since we want x, we take the square root of both sides we get

$x = \sqrt{y^{2} +5}$