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b) Find 3 different possible pairs of values for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] in this equation:
[tex]\[ 19 = ab + 7 \][/tex]
( [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are whole numbers. )

\begin{tabular}{|l|l|}
\hline
Value of [tex]\(a\)[/tex] & Value of [tex]\(b\)[/tex] \\
\hline
& \\
\hline
& \\
\hline
& \\
\hline
\end{tabular}


Sagot :

Certainly! Let's break down the problem step-by-step to find three different pairs of values for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that satisfy the equation:

[tex]\[ 19 = a \cdot b + 7 \][/tex]

First, we should rearrange the equation to make it easier to find pairs [tex]\((a, b)\)[/tex]:

[tex]\[ 19 - 7 = a \cdot b \implies 12 = a \cdot b \][/tex]

Now, we need to find pairs of whole numbers [tex]\((a, b)\)[/tex] such that their product is 12. Let's list the factor pairs of 12:

The factor pairs of 12 are:
1. [tex]\((1, 12)\)[/tex]
2. [tex]\((2, 6)\)[/tex]
3. [tex]\((3, 4)\)[/tex]
4. [tex]\((4, 3)\)[/tex]
5. [tex]\((6, 2)\)[/tex]
6. [tex]\((12, 1)\)[/tex]

From these pairs, we can select three different pairs where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are whole numbers:

[tex]\[ \begin{array}{|l|l|} \hline \text{Value of } a & \text{Value of } b \\ \hline 1 & 12 \\ \hline 2 & 6 \\ \hline 3 & 4 \\ \hline \end{array} \][/tex]

Thus, the three different possible pairs of values for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that satisfy the equation [tex]\(19 = a \cdot b + 7\)[/tex] are:

[tex]\[ (1, 12), (2, 6), (3, 4) \][/tex]