Complete the sums by filling in the missing numbers.

[tex]\[
\begin{array}{|c|c|c|c|c|}
\hline
J & \times & 4 & = & 20 \\
\hline
& \times & & & \\
\hline
& \times & & = & \\
\hline
= & & = & & = \\
\hline
15 & \times & & = & 60 \\
\hline
\end{array}
\][/tex]

[tex]\[
\begin{array}{|c|c|c|c|c|}
\hline
& \div & 4 & = & \\
\hline
& & & & \\
\hline
& \div & & = & 3 \\
\hline
= & & = & & = \\
\hline
& \div & 8 & = & 9 \\
\hline
\end{array}
\][/tex]



Answer :

Sure! Let's solve the given equations step-by-step to find the missing numbers.

### First Equation:
[tex]\[ J \times 4 = 20 \][/tex]
To find [tex]\( J \)[/tex], we divide both sides of the equation by 4:
[tex]\[ J = \frac{20}{4} = 5 \][/tex]

So, [tex]\( J = 5 \)[/tex].

### Second Equation:
[tex]\[ 15 \times X = 60 \][/tex]
To find [tex]\( X \)[/tex], we divide both sides of the equation by 15:
[tex]\[ X = \frac{60}{15} = 4 \][/tex]

So, [tex]\( X = 4 \)[/tex].

### Third Equation:
[tex]\[ Y \div 8 = 9 \][/tex]
To find [tex]\( Y \)[/tex], we multiply both sides of the equation by 8:
[tex]\[ Y = 9 \times 8 = 72 \][/tex]

So, [tex]\( Y = 72 \)[/tex].

### Fourth Equation:
[tex]\[ Z \div 4 = 3 \][/tex]
To find [tex]\( Z \)[/tex], we multiply both sides of the equation by 4:
[tex]\[ Z = 3 \times 4 = 12 \][/tex]

So, [tex]\( Z = 12 \)[/tex].

Therefore, the filled-in equations and their solutions are:
[tex]\[ \begin{pmatrix} 5 & \times & 4 & = & 20 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 15 & \times & 4 & = & 60 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 72 & \div & 8 & = & 9 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 12 & \div & 4 & = & 3 \\ \end{pmatrix} \][/tex]

The missing numbers are [tex]\( J = 5 \)[/tex], [tex]\( X = 4 \)[/tex], [tex]\( Y = 72 \)[/tex], and [tex]\( Z = 12 \)[/tex].