Complete the table and by inspection of the table find the solution to the equation.

$m + 10 = 16$

$\begin{matrix} m & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & \dots & \dots m + 10 & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{matrix}$

Open in App

Solution

$\begin{matrix} m & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 m + 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \end{matrix}$

The given equation is $m + 10 = 16$ .

Analysing the table above, we can understand that only $m = 6$ satisfies the given equation.

$∴ m = 6$ is the solution.

Suggest Corrections

Similar questions

Question 4(a)

Complete the table and by inspection of the table find the solution to the equation.

m+ 10 =16

$\begin{matrix} m & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & \dots & \dots m + 10 & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{matrix}$

Question 4(c)

Complete the table and by inspection of the table find the solution to the equation $\frac{z}{3} = 4.$

$\begin{matrix} z & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & \dots & \dots & \dots & \dots \frac{z}{3} & 2 \frac{2}{3} & 3 & 3 \frac{1}{3} & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{matrix}$

Question 4(b)

Complete the table and by inspection of the table find the solution to the equation 5t = 35.

$\begin{matrix} t & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & \dots & \dots & \dots & \dots & \dots 5 t & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{matrix}$

Question 4(d)

Complete the table and by inspection of the table find the solution to the equation m - 7 = 3.

$\begin{matrix} m & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & \dots & \dots m - 7 & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{matrix}$

Q. Question 22 (i)
(i) Complete the following table:

\begin{matrix} Event: Sum of 2 dice & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 Probability & \frac{1}{36} & \frac{5}{36} & \frac{1}{36} \end{matrix}

Join BYJU'S Learning Program

Complete the table and by inspection of the table find the solution to the equation. m+10=16 m12345678910⋯⋯⋯m+10⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯

m12345678910111213m+1011121314151617181920212223 The given equation is m+10=16. Analysing the table above, we can understand that only m=6 satisfies the given equation. ∴m=6 is the solution.