Replace the blank with an integer to make it a true statement.
$ \_\_\_\_\_ \times \left( { - 8} \right) = - 56 $
Answer
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Hint: In this problem, first we will assume that there is an integer $ x $ in the given blank. Then, we will use basic mathematical operations like multiplication and division. We know that multiplication of positive integers with negative integers will give negative results. We will use this basic knowledge to find required integers.
Complete step-by-step answer:
In this problem, we need to replace the blank with an integer such that the given statement $ \_\_\_\_\_ \times \left( { - 8} \right) = - 56 $ becomes true. Here the result is $ - 56 $ and it is a negative integer. On LHS we can see that there is an integer $ - 8 $ and it is also a negative integer. We know that multiplication of positive integers with negative integers will give negative results. According to this information, we can say that there must be positive integers in the blank. To find that positive integer, let us assume that $ x $ is an integer in the given blank. Therefore, we can write $ \left( x \right) \times \left( { - 8} \right) = - 56 \cdots \cdots \left( 1 \right) $ . Perform multiplication on LHS of the equation $ \left( 1 \right) $ , we get $ - 8x = - 56 \cdots \cdots \left( 2 \right) $ . By Cancelling the negative sign from both sides of the equation $ \left( 2 \right) $ , we get $ 8x = 56 \cdots \cdots \left( 3 \right) $ . We can see that the equation $ \left( 3 \right) $ is a linear equation in one variable $ x $ . To find the value of $ x $ , we will perform division on both sides of equation $ \left( 3 \right) $ . Let us divide by integer $ 8 $ on both sides of the equation $ \left( 3 \right) $ . Therefore, we get
$ \dfrac{{8x}}{8} = \dfrac{{56}}{8} \Rightarrow x = 7 $ . Therefore, we can say that if we replace the blank with an integer $ 7 $ then $ 7 \times \left( { - 8} \right) = - 56 $ becomes a true statement.
Note: This type of problem can be solved by using the basic knowledge of multiplication and division. Multiplication of two positive integers is always positive integer. Also multiplication of two negative integers is always positive integer. Multiplication of negative integers with positive integers will give negative integers.
Complete step-by-step answer:
In this problem, we need to replace the blank with an integer such that the given statement $ \_\_\_\_\_ \times \left( { - 8} \right) = - 56 $ becomes true. Here the result is $ - 56 $ and it is a negative integer. On LHS we can see that there is an integer $ - 8 $ and it is also a negative integer. We know that multiplication of positive integers with negative integers will give negative results. According to this information, we can say that there must be positive integers in the blank. To find that positive integer, let us assume that $ x $ is an integer in the given blank. Therefore, we can write $ \left( x \right) \times \left( { - 8} \right) = - 56 \cdots \cdots \left( 1 \right) $ . Perform multiplication on LHS of the equation $ \left( 1 \right) $ , we get $ - 8x = - 56 \cdots \cdots \left( 2 \right) $ . By Cancelling the negative sign from both sides of the equation $ \left( 2 \right) $ , we get $ 8x = 56 \cdots \cdots \left( 3 \right) $ . We can see that the equation $ \left( 3 \right) $ is a linear equation in one variable $ x $ . To find the value of $ x $ , we will perform division on both sides of equation $ \left( 3 \right) $ . Let us divide by integer $ 8 $ on both sides of the equation $ \left( 3 \right) $ . Therefore, we get
$ \dfrac{{8x}}{8} = \dfrac{{56}}{8} \Rightarrow x = 7 $ . Therefore, we can say that if we replace the blank with an integer $ 7 $ then $ 7 \times \left( { - 8} \right) = - 56 $ becomes a true statement.
Note: This type of problem can be solved by using the basic knowledge of multiplication and division. Multiplication of two positive integers is always positive integer. Also multiplication of two negative integers is always positive integer. Multiplication of negative integers with positive integers will give negative integers.
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