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**Hint:**We are provided with the ratio of two numbers and their sum to find the numbers. We will assume the common factor of both the numbers to be \[x\]. We will then form the equation based on the given conditions and then solve them to obtain the common factor. Using the common factor we will find the required numbers.

**Complete step-by-step answer:**Let us assume that the common factor of both our numbers is \[x\].

Now we are given that the two numbers are in the ratio of \[8:3\]. This means that after we divide both the numbers with each other and cancel out the common factor \[x\]from both the terms we get 8 in numerator and denominator. In other words, when we divide the numerator and denominator both with \[x\], we get 8 in the numerator and 3 in the denominator.

Let us find the first number now,

From the above concept, we can see that the first number would be the product of 8 and \[x\].

Thus, the first number \[ = 8x\]

Now let us find the second number,

From the above concept, we know that the second number would be the product of 3 and \[x\].

Thus, the second number \[ = 3x\]

We are given in the question that the sum of both our numbers is 143. Forming an equation based on this information, we get

\[8x + 3x = 143\]

Adding the like terms, we get

\[ \Rightarrow 11x = 143\]

Dividing both side by 11, we get

\[\begin{array}{l} \Rightarrow \dfrac{{11x}}{{11}} = \dfrac{{143}}{{11}}\\ \Rightarrow x = 13\end{array}\]

Thus, the common factor for both of our numbers is 13.

Now we will calculate both the numbers using the common factor.

The first number \[ = 8x = 8 \times 13 = 104\]

Thus, our first number is 104.

So, our second number will be the product of 3 and the common factor 13

The second number \[ = 3x = 3 \times 13 = 39\]

Thus, our second number is 39.

**\[\therefore\] The numbers are 104, and 39.**

**Note:**The alternate way to solve this question can be as follows –

Let us assume that the two numbers are \[x\] and \[y\] with \[x > y\].

Now we are given that the ratio of the two numbers is 8 : 3. This means that,

\[\dfrac{x}{y} = \dfrac{8}{3}\]

On solving the above equation, we get

\[3x = 8y\]

\[ \Rightarrow 3x - 8y = 0\]………..\[\left( 1 \right)\]

We are also given that the sum of the two numbers is 143.

\[x + y = 143\]……………….\[\left( 2 \right)\]

We will now solve equations \[\left( 1 \right)\] and \[\left( 2 \right)\] using the method of elimination.

Multiplying equation \[\left( 2 \right)\]by 8, we get,

\[8x + 8y = 143 \times 8\]

\[ \Rightarrow 8x + 8y = 1144\]…………\[\left( 3 \right)\]

Adding equation \[\left( 1 \right)\] and \[\left( 3 \right)\], we get

\[\begin{array}{l}3x - 8y + 8x + 8y = 0 + 1144\\ \Rightarrow 11x = 1144\end{array}\]

On dividing both sides of the equation by 11, we get

\[\begin{array}{l}\dfrac{{11x}}{{11}} = \dfrac{{1144}}{{11}}\\ \Rightarrow x = 104\end{array}\]

Thus, one number is 104. Let us back-substitute the value of \[x = 104\] in equation \[\left( 2 \right)\] and then solve it. On doing so we get,

\[\begin{array}{l}104 + y = 143\\ \Rightarrow y = 143 - 104\\ \Rightarrow y = 39\end{array}\]

Hence, another number is 39.

So, our two numbers are 104 and 39.

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