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Show that the product of a number and seven is equal to two more than the number.

The aim of the given question is to introduce word problems related to basic algebra and arithmetic operations.

To solve such questions we may need to first assume the required numbers as algebraic variables. Then we try to convert the given constraints into the form of algebraic equations. Finally, we solve these equations to find the values of the required numbers.

Expert Answer

Let $ x $ be the number that we want to find. Then:

\[ \text{ Product of } x \text{ and } 7 \ = \ ( x )( 7 ) \ = \ 7 x \]

And:

\[ \text{ Two more than } x \ = \ x \ + \ 2 \]

Under the given conditions and constraints, we can formulate the following equation:

\[ \text{ Product of } x \text{ and } 7 \ = \ \text{ Two more than } x \]

\[ \Rightarrow 7 x \ = \ x \ + \ 2 \]

Subtracting $ x $ from both sides:

\[ 7 x \ – \ x \ = \ x \ + \ 2 \ – \ x \]

\[ \Rightarrow 6 x \ = \ 2 \]

Dividing both sides by $ 6 $:

\[ \dfrac{ 1 }{ 6 } \times 6 x \ = \ \dfrac{ 1 }{ 6 } \times 2 \]

\[ \Rightarrow x \ = \ \dfrac{ 1 }{ 3 } \]

Which is the required number.

Numerical Result

\[ x \ = \ \dfrac{ 1 }{ 3 } \]

Example

Find two numbers such that the sum of both numbers is equal to 2 more than their product and one of the numbers is 2 more than the other number.

Let $ x $ and $ y $ be the number that we want to find. Then:

\[ \text{ Two more than product of } x \text{ and } y \ = \ ( x )( y ) \ + \ 2 \ = \ x y \]

\[ \text{ Sum of } x \text{ and } y \ = \ x \ + \ y  \ = \ \]

And:

\[ \text{ Two more than } x \ = \ x \ + \ 2 \]

Under the given conditions and constraints, we can formulate the following equations:

\[ \text{ Sum of } x \text{ and } y \ = \ \text{ Two more than product of } x \text{ and } y \]

\[ x \ + \ y \ = \ x y \ + \ 2 \ … \ … \ … \ ( 1 ) \]

And:

\[ x \ = \ y \ + \ 2 \ … \ … \ … \ ( 2 ) \]

Substituting the value of $ x $ from equation (2) in equation (1):

\[ ( y \ + \ 2 ) \ + \ y \ = \ ( y \ + \ 2 ) y \ + \ 2 \]

\[ \Rightarrow 2 y \ + \ 2 \ = \ y^2 \ + \ 2 y \ + \ 2 \]

Adding $ – 2 y – 2 $ on both sides:

\[ 2 y \ + \ 2 \ – \ 2 y \ – \ 2 = \ y^2 \ + \ 2 y \ + \ 2 \ – \ 2 y \ – 2 \]

\[ \Rightarrow 0 \ = \ y^2 \]

\[ \Rightarrow y \ = \ 0 \]

Substituting this value of $ y $ in equation (2):

\[ x \ = \ ( 0 ) \ + \ 2 \]

\[ \Rightarrow x \ = \ 2 \]

Hence, 0 and 2 are the required numbers.

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