Find the product using suitable properties.
$ 7\times \left( 50-2 \right) $
Answer
644.7k+ views
Hint: In order to solve this problem, we need to understand the distributive law of multiplication. The distributive property of multiplication over addition or subtraction can be used when a number is multiplied by a summation or subtraction of numbers. In the case of two brackets of multiplication, we can expand any bracket first. The answer will not change.
Complete step-by-step answer:
To solve this problem, we need to use the distributive property of multiplication
Let’s first understand what distributive law of multiplication means.
The distributive property of multiplication over addition or subtraction can be used when a number is multiplied by a summation or subtraction of numbers.
It is expressed mathematically as follows,
$ a\left( b+c \right)=ab+bc $
Where, a, b, c are just the constants.
In this problem, we can write 7 as 10 – 3.
Therefore, the expression becomes $ \left( 10-3 \right)\left( 50-2 \right) $ .
In this sum, we have two brackets with subtraction.
We need to open any bracket and perform distributive law.
Opening the first bracket we get,
$ \left( 10-3 \right)\left( 50-2 \right)=10\left( 50-2 \right)-3\left( 50-2 \right) $
Now, we can again use the distributive law of multiplication for each of the brackets,
We can simplify by,
$ 10\left( 50-2 \right)-3\left( 50-2 \right)=\left( 10\times 50 \right)+\left( 10\times -2 \right)+\left( -3\times 50 \right)+\left( -3\times -2 \right) $
Simplifying further we get,
$ \left( 10-3 \right)\left( 50-2 \right)=\left( 500 \right)+\left( -20 \right)+\left( -150 \right)+\left( 6 \right) $
Then the product of a negative and a positive number is always a negative number.
Therefore, $ \left( 10\times -2 \right)=-20 $ .
And the product of a negative and a negative number is always a positive number.
Therefore, $ \left( -3\times -2 \right)=+6 $ .
$ \left( 10-3 \right)\left( 50-2 \right)=336 $ .
Therefore, the product using the distributive law of multiplication is 336.
Note: We will get the same answer expanding the second bracket first. We can show that as follows,
\[\begin{align}
& \left( 10-3 \right)\left( 50-2 \right)=50\left( 10-3 \right)-2\left( 10-3 \right) \\
& =\left( 500\times 10 \right)+\left( 50\times -3 \right)+\left( -2\times 10 \right)+\left( -2\times -3 \right) \\
& =500-150-20+6 \\
Complete step-by-step answer:
To solve this problem, we need to use the distributive property of multiplication
Let’s first understand what distributive law of multiplication means.
The distributive property of multiplication over addition or subtraction can be used when a number is multiplied by a summation or subtraction of numbers.
It is expressed mathematically as follows,
$ a\left( b+c \right)=ab+bc $
Where, a, b, c are just the constants.
In this problem, we can write 7 as 10 – 3.
Therefore, the expression becomes $ \left( 10-3 \right)\left( 50-2 \right) $ .
In this sum, we have two brackets with subtraction.
We need to open any bracket and perform distributive law.
Opening the first bracket we get,
$ \left( 10-3 \right)\left( 50-2 \right)=10\left( 50-2 \right)-3\left( 50-2 \right) $
Now, we can again use the distributive law of multiplication for each of the brackets,
We can simplify by,
$ 10\left( 50-2 \right)-3\left( 50-2 \right)=\left( 10\times 50 \right)+\left( 10\times -2 \right)+\left( -3\times 50 \right)+\left( -3\times -2 \right) $
Simplifying further we get,
$ \left( 10-3 \right)\left( 50-2 \right)=\left( 500 \right)+\left( -20 \right)+\left( -150 \right)+\left( 6 \right) $
Then the product of a negative and a positive number is always a negative number.
Therefore, $ \left( 10\times -2 \right)=-20 $ .
And the product of a negative and a negative number is always a positive number.
Therefore, $ \left( -3\times -2 \right)=+6 $ .
$ \left( 10-3 \right)\left( 50-2 \right)=336 $ .
Therefore, the product using the distributive law of multiplication is 336.
Note: We will get the same answer expanding the second bracket first. We can show that as follows,
\[\begin{align}
& \left( 10-3 \right)\left( 50-2 \right)=50\left( 10-3 \right)-2\left( 10-3 \right) \\
& =\left( 500\times 10 \right)+\left( 50\times -3 \right)+\left( -2\times 10 \right)+\left( -2\times -3 \right) \\
& =500-150-20+6 \\
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