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\[2x+5=9\] 1 Why did we take this step? Because we have \(2x+5\) 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.\[2x=9-5\] 2 Simplify \(9-5\) to \(4\). \[2x=4\] 3 Why did we take this step? Because we have \(2x\) on the left side, and we want only \(x\). Therefore, we divide to undo the multiplication.\[x=\frac{4}{2}\] 4 Simplify \(\frac{4}{2}\) to \(2\). \[x=2\] Done \[3(3-2x)=5(7-5x)\] 1 Why did we take this step? Because by expanding, we distribute the terms and remove the parentheses , which usually allows us to simplify the expression further.\[9-6x=35-25x\] 2 Why did we take this step? Because this helps us cancel \(9\). Since our goal is to solve for \(x\), canceling any term that is not \(x\) is helpful .\[-6x=35-25x-9\] 3 Simplify \(35-25x-9\) to \(-25x+26\). \[-6x=-25x+26\] 4 Why did we take this step? Because in the previous step, \(x\) is on both sides of the equation. Since our goal is to solve for \(x\), we need it on one side only .\[-6x+25x=26\] 5 Simplify \(-6x+25x\) to \(19x\). \[19x=26\] 6 Why did we take this step? Because we have \(19x\) on the left side, and we want only \(x\). Therefore, we divide to undo the multiplication.\[x=\frac{26}{19}\] Done Decimal Form: 1.368421 \[6x=12\] 1 Why did we take this step? Because we have \(6x\) on the left side, and we want only \(x\). Therefore, we divide to undo the multiplication.\[x=\frac{12}{6}\] 2 Simplify \(\frac{12}{6}\) to \(2\). \[x=2\] Done \[\sqrt{x+4}=x+5\] 1 Square both sides. \[x+4={x}^{2}+10x+25\] 2 Move all terms to one side. \[x+4-{x}^{2}-10x-25=0\] 3 Simplify \(x+4-{x}^{2}-10x-25\) to \(-9x-21-{x}^{2}\). \[-9x-21-{x}^{2}=0\] 4 Use the Quadratic Formula. \[x=\frac{9+\sqrt{3}\imath }{-2},\frac{9-\sqrt{3}\imath }{-2}\] 5 Simplify solutions. \[x=-\frac{9+\sqrt{3}\imath }{2},-\frac{9-\sqrt{3}\imath }{2}\] Done \[{x}^{4}+9{x}^{3}+9{x}^{2}-85x-150\] 1 Factor \({x}^{4}+9{x}^{3}+9{x}^{2}-85x-150\) using Polynomial Division. \[({x}^{3}+7{x}^{2}-5x-75)(x+2)\] 2 Factor \({x}^{3}+7{x}^{2}-5x-75\) using Polynomial Division. \[({x}^{2}+10x+25)(x-3)(x+2)\] 3 Why did we take this step? Because \({a}^{2}+2ab+{b}^{2}\) is a common expression with a known factored form. This allows us to factor the expression in the next step.\[({x}^{2}+2(x)(5)+{5}^{2})(x-3)(x+2)\] 4 Use Square of Sum: \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\). \[{(x+5)}^{2}(x-3)(x+2)\] Done \[{w}^{2}+8w-65\] 1 Ask: Which two numbers add up to \(8\) and multiply to \(-65\)? \(-5\) and \(13\) 2 Rewrite the expression using the above. \[(w-5)(w+13)\] Done |