Question

Match the following equations with their respective values of $x$, if $y$ = 15.

• A. 17.4
• B. 12.4
• C. 9.8

Answer 1

Given:
$x^2=10y+4$ ---------(1) and $y=15$

Substitute the value of $y$ in (1)

We get:

$x^2=10\left(15\right)+4$

$⟹$ $x^2=150+4$

$⟹$ $x^2=154$

$⟹$ $\sqrt{x^2}=\sqrt{154}$

$⟹$ $x=\sqrt{154}$

$⟹$ $144<154<169$

$⟹$ $12<\sqrt{154}<13$

154 is closer to 144 than 169. Now, starting from 12 and increasing by a step of one-tenth till we reach closer to 154, we get:

${12.4}^2=153.76$ and ${12.5}^2=156.25$

153.76 is closer to 154 than 156.25.

$\therefore x=12.4$


Given
$x^2=22y-26$ ---------(2) and $y=15$

Substitute the value of $y$ in (2)

We get:

$x^2=22\left(15\right)-26$

$x^2=330-26$

$x^2=304$

$\sqrt{x^2}=\sqrt{304}$

$x=\sqrt{304}$

304 lies between the 289 and 324, which are prefect squares of 17 and 18 respectively

$289<304<324$

$17<\sqrt{304}<18$

304 is closer to 289 than 324. Now, startting from 17 and increasing by a step of one tenth until you reach closer to 304, we get:

${17.4}^2=302.76\text{and}{17.5}^2=306.25$

302.76 is closer to 304 than 306.25

$\therefore x=17.4$


Given
$x^2=142-3y$ ---------(3) and $y$ = 15

Substitute the value of y in equation (3)

We get:

$x^2=142-3\left(15\right)$

$x^2=142-45$

$x^2=97$

$\sqrt{x^2}=\sqrt{397}$

$x=\sqrt{97}$

97 lies between the 81 and 100, which are prefect squares of 12 and 13 respectively

$81<97<100$

$9<\sqrt{97}<10$

97 is closer to 100 than 81. Now, starting from 10 and decreasing by a step of one-tenth till we reach closer to 97, we get:

${9.8}^2=96.04\text{and}{9.9}^2=98.01$

96.04 is closer to 97 than 98.01.

$\therefore x=9.8$