How do you simplify \[5\left( {x - 3} \right)\]?
Answer
624.9k+ views
Hint: In the given question, we have been given an algebraic expression. In the expression, a constant is multiplied with a bracket. The bracket consists of two terms – one variable and one constant, separated by a minus sign. We have to solve the expression. To do that, we first multiply the constant outside the bracket with the first term (the variable) then we keep the minus sign and then we multiply it with the other constant, and that gives us the answer.
Formula Used:
\[a\left( {b - c} \right) = a \times b - a \times c\]
Complete step by step answer:
The given expression to be simplified is:
\[5\left( {x - 3} \right)\]
We know, \[a\left( {b - c} \right) = a \times b - a \times c\].
Substituting \[a = 5\], \[b = x\] and \[c = 3\], we have,
\[5\left( {x - 3} \right) = 5x - 15\]
Note:
In the given question, we had to simplify the value of an expression. To do that, we multiply the term outside the bracket with the terms inside the bracket sequentially, keep the sign between the terms and we have got the answer. It is an important point to remember that if the term outside the bracket is negative, it changes the sign inside the bracket (positive becomes negative and vice-versa) and any positive term becomes negative.
Formula Used:
\[a\left( {b - c} \right) = a \times b - a \times c\]
Complete step by step answer:
The given expression to be simplified is:
\[5\left( {x - 3} \right)\]
We know, \[a\left( {b - c} \right) = a \times b - a \times c\].
Substituting \[a = 5\], \[b = x\] and \[c = 3\], we have,
\[5\left( {x - 3} \right) = 5x - 15\]
Note:
In the given question, we had to simplify the value of an expression. To do that, we multiply the term outside the bracket with the terms inside the bracket sequentially, keep the sign between the terms and we have got the answer. It is an important point to remember that if the term outside the bracket is negative, it changes the sign inside the bracket (positive becomes negative and vice-versa) and any positive term becomes negative.
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