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What number must be added to each of the numbers 8, 13, 18, and 28 so that the four resulting numbers are in proportion?

Let [tex]\( x \)[/tex] be the number which, when added to each of the numbers 8, 13, 18, and 28, gives four numbers that are in proportion.

The four new numbers obtained by adding [tex]\( x \)[/tex] to the given numbers are:
[tex]\[ 8 + x, 13 + x, 18 + x, \text{and} 28 + x \][/tex]

Determine the value of [tex]\( x \)[/tex].



Answer :

GinnyAnswer
To find the number [tex]\( x \)[/tex] that must be added to each of the numbers 8, 13, 18, and 28 so that the four resulting numbers are in proportion, let's define the resulting numbers as follows:

- [tex]\( 8 + x \)[/tex]
- [tex]\( 13 + x \)[/tex]
- [tex]\( 18 + x \)[/tex]
- [tex]\( 28 + x \)[/tex]

These four numbers will be in proportion if:

[tex]\[ \frac{8 + x}{13 + x} = \frac{18 + x}{28 + x} \][/tex]

1. Start by writing the proportion:
[tex]\[ \frac{8 + x}{13 + x} = \frac{18 + x}{28 + x} \][/tex]

2. Cross-multiply to form the equation:
[tex]\[ (8 + x)(28 + x) = (13 + x)(18 + x) \][/tex]

3. Expand both sides of the equation:
[tex]\[ (8 + x)(28 + x) = 8 \cdot 28 + 8 \cdot x + x \cdot 28 + x^2 = 224 + 8x + 28x + x^2 = 224 + 36x + x^2 \][/tex]
[tex]\[ (13 + x)(18 + x) = 13 \cdot 18 + 13 \cdot x + x \cdot 18 + x^2 = 234 + 13x + 18x + x^2 = 234 + 31x + x^2 \][/tex]

4. Set the expanded forms equal to each other:
[tex]\[ 224 + 36x + x^2 = 234 + 31x + x^2 \][/tex]

5. Simplify the equation by subtracting [tex]\(x^2\)[/tex] from both sides (they cancel out), and then subtracting 224 and 31x from both sides:
[tex]\[ 36x - 31x = 234 - 224 \][/tex]
[tex]\[ 5x = 10 \][/tex]

6. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{10}{5} = 2 \][/tex]

Hence, the number which must be added to each of the numbers 8, 13, 18, and 28 to make the resulting numbers proportional is [tex]\( x = 2 \)[/tex].

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