Solve for x : x 2 x + 1 + 2 x + 1 x = 2 16 21

Converting the given mixed fraction into improper fraction, we get 2 16 21 = 2 × 21 + 16 21 = 58 21 . Now, x 2 x + 1 + 2 x + 1 x = 2 16 21 ⇒ x 2 x + 1 + 2 x + 1 x = 58 21 ⇒ x × x + 2 x + 1 × 2 x + 1 x 2 x + 1 = 58 21 ⇒ x 2 + 2 x + 1 2 2 x × x + x = 58 21 ⇒ x 2 + 4 x 2 + 1 + 4 x 2 x 2 + x = 58 21 ⇒ 21 × 5 x 2 + 4 x + 1 = 58 2 x 2 + x ⇒ 21 × 5 x 2 + 21 × 4 x + 21 = 58 × 2 x 2 + 58 x ⇒ 105 x 2 + 84 x + 21 = 116 x 2 + 58 x ⇒ 116 x 2 - 105 x 2 + 58 x - 84 x - 21 = 0 ⇒ 11 x 2 - 26 x - 21 = 0 ⇒ 11 x 2 - 33 x + 7 x - 21 = 0 ⇒ 11 x x - 3 + 7 x - 3 = 0 ⇒ 11 x + 7 x - 3 = 0 ⇒ x = - 7 11 , 3 Therefore the value of x can be - 7 11 or 3 .

Solve for x : 1 x + 5 + 1 x = 1 x - 4

We have, 1 x + 5 + 1 x = 1 x - 4 ⇒ x + x + 5 x x + 5 = 1 x - 4 ⇒ 2 x + 5 x × x + 5 x = 1 x - 4 ⇒ 2 x + 5 x 2 + 5 x = 1 x - 4 ⇒ 2 x + 5 x - 4 = x 2 + 5 x ⇒ 2 x × x - 2 x × 4 + 5 x - 20 = x 2 + 5 x ⇒ 2 x 2 - 8 x + 5 x - 20 = x 2 + 5 x ⇒ 2 x 2 - x 2 - 8 x + 5 x - 20 - 5 x = 0 ⇒ x 2 - 8 x - 20 = 0 ⇒ x 2 + 2 x - 10 x - 20 = 0 ⇒ x x + 2 - 10 x + 2 = 0 ⇒ x + 2 x - 10 = 0 Therefore the value of x can be - 2 or 10 .

Given below is the equation that convert Fahrenheit ( F ) to Celsius ( c ) , c = 5 F - 160 9 . If the temperature is 86 ° F , what is the temperature in Celsius?

The equation that converts Fahrenheit to Celsius is: c = 5 F - 160 9 . Given that temperature in Fahrenheit is 86 ° F . Now c = 5 F - 160 9 = 5 × 86 - 160 9 = 430 - 160 9 = 270 9 = 30 Therefore the temperature in Celsius is 30 ° C .

Given below is the equation that convert Fahrenheit ( F ) to Celsius ( c ) , c = 5 F - 160 8 . If the temperature is 35 ° C , what is the temperature in Fahrenheit?

The equation that shows conversion between Fahrenheit and Celsius is: c = 5 F - 160 9 . It can be written as: 9 c = 5 F - 160 ⇒ 5 F = 9 c + 160 ⇒ F = 9 c + 160 5 Given that temperature in Celsius is 35 ° C . Now, F = 9 c + 160 5 = 9 × 35 + 160 5 = 315 + 160 5 F = 475 5 = 95 The temperature in Fahrenheit is 95 ° F .

Find 3 points which satisfy the Linear equation x 2 - y 3 = 1 . What type of graph we obtain if we join these points (without plotting the graph).

The given linear equation is x 2 - y 3 = 1 . For x = 0 , 0 2 - y 3 = 1 ⇒ - y = 3 ⇒ y = - 3 . So one of the points which satisfy the given equation is 0 , - 3 . For y = 0 , x 2 - 0 3 = 1 ⇒ x = 2 . Another point obtained is 2 , 0 . For y = 3 , x 2 - 3 3 = 1 ⇒ x 2 = 1 + 1 ⇒ x = 2 × 2 = 4 . The third point which satisfies the given equation is 4 , 3 . We know that if the given linear equation is represented geometrically on a graph, we obtain a straight line. Therefore, if we join these points, the type of graph obtained is straight line.