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\[6\div 2(1+2)\] 1 Why did we take this step? Because of PEMDAS (the order of operations), we ask the questions below in order.Any parentheses ? Yes. Any exponents ? --Any multiplication / division ? --Any addition / subtraction ? --simplify terms in parentheses first. In other words, we simplify \(1+2\).\[6\div 2\times 3\] 2 Why did we take this step? Because of PEMDAS (the order of operations), we ask the questions below in order.Any parentheses ? No.Any exponents ? No.Any multiplication / division ? Yes, division. Any addition / subtraction ? --divide first. In other words, we simplify \(6\div 2\).\[3\times 3\] 3 Simplify. \[9\] Done \[\frac{2+x}{3}=8\] 1 Why did we take this step? Because we have \(\frac{2+x}{3}\) on the left side, and we want only \(x\). Using Reverse PEMDAS, we ask the questions below in order.Any addition / subtraction outside parentheses? No.Any multiplication / division outside parentheses? Yes, division. Any exponents ? --Any parentheses ? --Therefore, we multiply to undo the division.\[2+x=8\times 3\] 2 Simplify \(8\times 3\) to \(24\). \[2+x=24\] 3 Why did we take this step? Because we have \(2+x\) on the left side, and we want only \(x\). Therefore, we subtract to undo the addition.\[x=24-2\] 4 Simplify \(24-2\) to \(22\). \[x=22\] Done \[3x+7=5\] 1 Why did we take this step? Because we have \(3x+7\) on the left side, and we want only \(x\). Using Reverse PEMDAS, we ask the questions below in order.Any addition / subtraction outside parentheses? Yes, addition. Any multiplication / division outside parentheses? --Any exponents ? --Any parentheses ? --Therefore, we subtract to undo the addition.\[3x=5-7\] 2 Simplify \(5-7\) to \(-2\). \[3x=-2\] 3 Why did we take this step? Because we have \(3x\) on the left side, and we want only \(x\). Therefore, we divide to undo the multiplication.\[x=-\frac{2}{3}\] Done Decimal Form: -0.666667 \[{x}^{2}{x}^{3}{y}^{5}{y}^{4}\] 1 Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\). Use Why did we take this step? Because the Product Rule simplifies the expression. Let us take \({x}^{2}{x}^{3}\) as an example. You can think of \({x}^{2}\) as 2 copies of \(x\), and \({x}^{3}\) as 3 copies of \(x\). Therefore:In this example, we end up with 5 copies of \(x\) in total, which is \({x}^{5}\).\[{x}^{2+3}{y}^{5+4}\] 2 Simplify \(2+3\) to \(5\). \[{x}^{5}{y}^{5+4}\] 3 Simplify \(5+4\) to \(9\). \[{x}^{5}{y}^{9}\] Done \[{x}^{4}-36\] 1 Why did we take this step? Because \({a}^{2}-{b}^{2}\) is a common expression with a known factored form. This allows us to factor the expression in the next step.\[{({x}^{2})}^{2}-{6}^{2}\] 2 Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\). \[({x}^{2}+6)({x}^{2}-6)\] Done |