Question 1 Solve Equations with Decimal Coefficients. In the following exercises, solve the equation by clearing the decimals. 0.10 𝑑 + 0.25 ( 𝑑 + 7 ) = 5.25 0.10d+0.25(d+7)=5.25 Answer 1 Step 1 The given equation is 0.10 𝑑 + 0.25 ( 𝑑 + 7 ) = 5.25 0.10d+0.25(d+7)=5.25. To clear the decimals, we multiply each term by 100. Step 2 100 × 0.10 𝑑 + 100 × 0.25 ( 𝑑 + 7 ) = 100 × 5.25 100×0.10d+100×0.25(d+7)=100×5.25 10 𝑑 + 25 ( 𝑑 + 7 ) = 525 10d+25(d+7)=525 Expanding, 10 𝑑 + 25 𝑑 + 175 = 525 10d+25d+175=525 Combine like terms, 35 𝑑 = 350 35d=350 Divide by 35, 𝑑 = 10 d=10 Therefore, 𝑑 = 10 d=10 is the solution of 0.10 𝑑 + 0.25 ( 𝑑 + 7 ) = 5.25 0.10d+0.25(d+7)=5.25. Question 2 Solve with Decimal Coefficients. In the following exercises, solve the equation by clearing the decimals. 0.7 𝑥 + 0.4 = 0.6 𝑥 + 2.4 0.7x+0.4=0.6x+2.4 Answer 1 Step 1 The given equation is 0.7 𝑥 + 0.4 = 0.6 𝑥 + 2.4 0.7x+0.4=0.6x+2.4. To clear the decimals, we multiply each term by 10. Step 2 10 × 0.7 𝑥 + 10 × 0.4 = 10 × 0.6 𝑥 + 10 × 2.4 10×0.7x+10×0.4=10×0.6x+10×2.4 7 𝑥 + 4 = 6 𝑥 + 24 7x+4=6x+24 Transfer the terms to get similar terms together, 7 𝑥 − 6 𝑥 = 24 − 4 7x−6x=24−4 𝑥 = 20 x=20 Therefore, 𝑥 = 20 x=20 is the solution of 0.7 𝑥 + 0.4 = 0.6 𝑥 + 2.4 0.7x+0.4=0.6x+2.4. Question 3 Exponential and Logarithmic Equations. Solve the equation. Find the exact solution if possible; otherwise, use a calculator to approximate to two decimals. log ⁡ 8 ( 𝑥 + 5 ) − log ⁡ 8 ( 𝑥 − 2 ) = 1 log 8 (x+5)−log 8 (x−2)=1 Answer 1 Step 1 The given logarithmic equation is log ⁡ 8 ( 𝑥 + 5 ) − log ⁡ 8 ( 𝑥 − 2 ) = 1 log 8 (x+5)−log 8 (x−2)=1 Using the law of logarithms for the difference of two logarithms, log ⁡ 8 ( 𝑥 + 5 𝑥 − 2 ) = 1 log 8 ( x−2 x+5 )=1 Rewrite in exponential form, 𝑥 + 5 𝑥 − 2 = 8 1 x−2 x+5 =8 1 𝑥 + 5 𝑥 − 2 = 8 x−2 x+5 =8 Cross-multiply, 𝑥 + 5 = 8 ( 𝑥 − 2 ) x+5=8(x−2) 𝑥 + 5 = 8 𝑥 − 16 x+5=8x−16 − 7 𝑥 = − 21 −7x=−21 𝑥 = 3 x=3 Therefore, 𝑥 = 3 x=3 is the solution of log ⁡ 8 ( 𝑥 + 5 ) − log ⁡ 8 ( 𝑥 − 2 ) = 1 log 8 (x+5)−log 8 (x−2)=1. Question 4 Solving Inequalities Graphically: Use a graphing device to solve the inequality. Express your answer using interval notation, with the endpoints of the interval rounded to two decimals. 𝑥 4 + 𝑥 3 ≤ 5 𝑥 2 + 4 𝑥 − 5 x 4 +x 3 ≤5x 2 +4x−5 Answer 1 Step 1 Consider the following inequality, 𝑥 4 + 𝑥 3 ≤ 5 𝑥 2 + 4 𝑥 − 5 x 4 +x 3 ≤5x 2 +4x−5 Subtract both sides to get the following, 𝑥 4 + 𝑥 3 − 5 𝑥 2 − 4 𝑥 + 5 ≤ 0 x 4 +x 3 −5x 2 −4x+5≤0 Using graphing software, the solutions for 𝑥 ≈ 0.74 x≈0.74 and 𝑥 ≈ 1.95 x≈1.95 were obtained. Graphing, we find the interval where the inequality holds. From the graph, 𝑥 ∈ [ 0.74 , 1.95 ] x∈[0.74,1.95] So, the solution is 𝑥 ∈ [ 0.74 , 1.95 ] x∈[0.74,1.95]. Question 5 Solving Equations and Inequalities Graphically. Solve the given equation or inequality graphically. State your answers rounded to two decimals. a) 5 𝑥 2 − 𝑥 3 = − 𝑥 2 + 3 𝑥 + 4 5x 2 −x 3 =−x 2 +3x+4 b) 5 𝑥 2 − 𝑥 3 ≤ − 𝑥 2 + 3 𝑥 + 4 5x 2 −x 3 ≤−x 2 +3x+4 Answer 1 Step 1 To solve 5 𝑥 2 − 𝑥 3 = − 𝑥 2 + 3 𝑥 + 4 5x 2 −x 3 =−x 2 +3x+4 graphically, set 𝑓 ( 𝑥 ) = 5 𝑥 2 − 𝑥 3 and 𝑔 ( 𝑥 ) = − 𝑥 2 + 3 𝑥 + 4 f(x)=5x 2 −x 3 andg(x)=−x 2 +3x+4 Find the points of intersection by graphing 𝑓 ( 𝑥 ) f(x) and 𝑔 ( 𝑥 ) g(x). Using a graphing tool, find where the graphs intersect to get the solutions for part (a). For part (b), determine where 𝑓 ( 𝑥 ) ≤ 𝑔 ( 𝑥 ) f(x)≤g(x) by shading the region on the graph. (The specific points of intersection or shaded regions will be identified using the graphing tool.)