Questions and Answers #9 Equations, Expressions and Inequalities

Questions and Answers Sheet 9 Equations, Expressions and Inequalities Question #1 To calculate: The solution of the equation 𝑥 2 − 5 = (𝑥 + 2)(𝑥 − 4). Answer: Formula used: The distributive property is a(b+c)=ab+ac . Calculation: Consider the equation 𝑥 2 − 5 = (𝑥 + 2)(𝑥 − 4) . Apply the distributive property. 𝑥 2 − 5 = 𝑥 2 − 4𝑥 + 2𝑥 − 8 𝑥 2 − 5 = 𝑥 2 − 2𝑥 −8 -5=-2x-8 8-5=-2x Further solve the equation. 3=-2x 3 𝑥=− 2 Check the solution 𝑥 = − 3 2 in the equation 𝑥 2 − 5 = (𝑥 + 2)(𝑥 − 4). ? 3 3 (− ) 2 − 5 ={(− ) + 2} 2 2 ? −3 + 4 −3 − 8 9 − 20 3 9 {(− ) − 4} − 5 =( )( ) 2 4 2 2 4 ? 1 11 =( )(− ) 2 2 − 11 4 =− 11 4 Thus, this solution is true. Therefore, thesolution of the equation 𝑥 2 − 5 = (𝑥 + 2)(𝑥 − 4) is 3. Question #2 To calculate: The equation 3𝑦 + 2[5(𝑦 − 4) − 2] = 5𝑦 + 6(7 + 𝑦) − 3 Answer: Calculation: Consider the given equation, 3𝑦 + 2[5(𝑦 − 4) − 2] = 5𝑦 + 6(7 + 𝑦) − 3 Now, apply the distributive property: 3𝑦 + 2[5(𝑦 − 4) − 2𝑟𝑖𝑔ℎ𝑡] = 5𝑦 + 6(7 + 9) − 3 3𝑦 + 2[5𝑦— 20 − 2] = 5𝑦 + 42 + 6𝑦— 3 3𝑦 + 10𝑦— 44 = 5𝑦 + 42 + 6𝑦— 3 Combine the like terms: 3𝑦 + 10𝑦— 44 = 5𝑦 + 42 + 6𝑦— 3 13𝑦 − 44 = 11𝑦 + 39 Now, subtract I ly from both sides: 13𝑦 − 44 = 11𝑦 + 39 13𝑦— 44— 11𝑦 = 11𝑦 + 39 + 11𝑦 Combine the like terms: 13𝑦— 44— 11𝑦 = 11𝑦 + 39 + 11𝑦 2𝑦 − 44 = 39 Add 44 to both sides: 2y-44+44=39+44 Combine the like terms: 2y=83 Divide both sides by 2: 83 𝑦= 2 Now, check the solution by put the value of y = © in the equation 3𝑦 + 2[5(𝑦 − 4)— 2] = 5𝑦 + 6(7 + 9𝑟𝑖𝑔ℎ𝑡) − 3. 3𝑦 + 2[5(𝑦 − 4) − 2] = 5𝑦 + 6(7 + 𝑦) − 3 83 83 ) + 2[5( − 4) − 2] 2 2 ? 83 83 249 415 = 5( ) + 6(7 + ) − 3( ) + 2[( − 20) − 2] 2 2 2 2 ? 415 498 249 830 =( ) + (42 + ) − 3( ) +[ − 40 − 4] 2 2 2 2 ? 415 498 =( ) + 42 + −3 2 2 3( On further solution: 249 830 + − 44 2 2 ? 415 498 249 + 830 − 88 = + + 39 2 2 2 ? 415 + 498 + 78 991 = 2 2 ? 991 = 2 Hence, the solution set is 𝑦 = 83 2 . Question #3 To calculate: The solution of the equation 3𝑦 2 − 4𝑦 = 8 − 6𝑦 Answer: Consider the equation, 3𝑦 2 − 4𝑦 = 8 − 6𝑦 Set one side of the equation equal to zero. 3𝑦 2 − 4𝑦 − (8 − 6𝑦) = 0 3𝑦 2 − 4𝑦 − 8 + 6𝑦 = 0 3𝑦 2 + 2𝑦 − 8 = 0𝑦 Make the factors of the equation. 3𝑦 2 + (6 − 4)𝑦 − 8 = 0 (3𝑦 2 + 6𝑦) + (−4𝑦 − 8) = 0 3𝑦(𝑦 + 2) − 4(𝑦 + 2) = 0𝑦 (3𝑦 − 4)(𝑦 + 2) = 0 Use the zero-product property. 3y-4=0 or y+2=0 3y=4 or y=-2 4 𝑦= 3 4 Check the solution 𝑦 = in the equation 3𝑦 2 − 4𝑦 = 8 − 6𝑦. 3 4 4 3( ) 2 − 4( ) 3 3 ? 4 16 16 = 8 − 6( )3( ) − 3 9 3 ? 24 16 16 =8 − − 3 3 3 ? 24 − 24 = 3 0=0 So, the answer is verified. Check the solution y=-2 in the equation 3𝑦 2 − 4𝑦 = 8 − 6𝑦. 3(−2)2 − 4(−2) ? = 8 − 6(−2)3(4) + 8 ? ? = 8 + 1212 + 8 = 20 20=20 So, the answer is verified. 4 Therefore, thesolution of the equation 3𝑦 2 − 4𝑦 = 8 − 6𝑦 is { , −2}. 3 Question #4 To calculate: The pounds of peanuts that must be added to an existing mixture of 45% peanuts to make 201b of a mixture that is 56% peanuts when at a candy and nut shop. Answer: Consider x represent the number of pounds of 100% peanuts and y represent the number of pounds of 45% peanuts. It is provided that some pounds of peanuts must be added to an existing mixture of 45% peanuts to make 201b of a mixture that is 56% peanuts. Construct the following table on the basis of provided information: 100% Peanuts 45% Peanuts 56% Peanuts Amountofmixture x y 20 Peanuts x 0.45 y 0.56(20) = 11.2 So, the equations obtained in two variables are: x+y=20 x+0.45y=112 Multiply the second equation with 100 to clear decimals: x+y=20.........(1) And, 100x+45y=1120......(2) Multiply with -100 in equation (1) -100x—100y=-2000 Add equation (1) and (2) -100x-100y=-2000 100x+45y=1120 −55 y = −880 Simplify the above equation, -55y=-880 y=16 Now, substitute 16 for y in equation (1) and solve for x x+(16)=20 (Subtracting 16 from both sides) x=4 Therefore, the manager should mix 41b of peanuts with 161b of the 45% mixture. Question #5 To calculate: The partial decomposition of 𝑥 2 −2𝑥−21 𝑥 3 +7𝑥 Answer: Calculation: Consider the provided expression: 𝑥 2 −2𝑥−21 𝑥 3 +7𝑥 Here, 𝑓(𝑥) = 𝑥? −2𝑥 − 21 and 𝑔(𝑥) = 𝑥 ∘ + 𝑇𝑥 Factorize g(x): 𝑔(𝑥) = 𝑥 3 + 7𝑥 = 𝑥(𝑥 2 + 7) So, the expression can be written as: 𝑥 2 −2𝑥−21 𝑥 3 +7𝑥 = 𝑥 2 −2𝑥−21 𝑥(𝑥 3 +7𝑥) Here, in g (x), there are two factors, one linear and one quadratic. So, the expression can be decomposed as: 𝑥 2 −2𝑥−21 𝑥(𝑥 3 +7𝑥) 𝐴 𝐵𝑥+𝐶 𝑥 𝑥 2 +7 = + Here, Least Common Divisor is 𝑥(𝑥 2 + 7) Multiply both sides of (1) by the Least Common Divisor to clear fractions: 𝑥(𝑥 2 + 7) [ 𝑥 2 −2𝑥−21 𝑥(𝑥 2 +7𝑥) −𝐴 ] = 𝑥(𝑥 2 + 7) [ 𝑥 + 𝐵𝑥+𝐶 𝑥 2 +7𝑥 ] simplify to obtain: 𝑥 2 + 2𝑥 − 21 = 𝐴(𝑥 2 + 7) + (𝐵𝑥 + 𝐶)𝑥 𝑥 2 + 2𝑥 − 21 = 𝐴𝑥 2 + 7 + 𝐵𝑥 2 + 𝐶𝑥 𝑥 2 + 2𝑥 − 21 = (𝐴 + 𝐶)𝑥 2 + 𝐶𝑥 + 7𝐴 Compare the coefficients of x,x? and constant terms: A+B=1, C=-2 and 7A=-21 Solve for A the equation 7A=-21 7A=-21 A=-3 Substitute -3 for A in A+B=1 and simplify for B: (-3)+B=1 B=4 Substitute the obtained values of A, B and Cin (1): 𝑥 2 −2𝑥−21 𝑥(𝑥 3 +7𝑥) 3 4𝑥−2 𝑥 𝑥 2 +7 = + Therefore, the partial fraction decomposition for 𝑥 2 −2𝑥−21 𝑥(𝑥 3 +7𝑥) 3 4𝑥−2 𝑥 𝑥 2 +7 is + Question #6 To calculate: The partial decomposition of the function −12𝑥−29 2𝑥 2 +11𝑥+15 . Answer: Given Information: The provided expression is −12𝑥−29 2𝑥 2 +11𝑥+15 . Formula used: Decomposition of 𝑓(𝑥) 𝑔(𝑥) into Partial Fractions: Consider a rational expression 𝑓(𝑥) 𝑔(𝑥) , where f(x) and g(x) are polynomial with real coefficients, 𝑔(𝑥) ≠ 0, and the degree of f(x) is less than degree of g(x). Step 1: Factor the denominator g(x) completely into linear factors of the form (𝑎𝑥 + 𝑏)𝑚 and quadratic factors of the form (𝑎𝑥 2 + 𝑏𝑥 + 𝑐)𝑛 that are not further factorable over the integers. Step 2: Set up the form of decomposition. That is, write the original rational expression 𝑓(𝑥) 𝑔(𝑥) as a sum of simpler fractions using these guidelines. Note that 𝐴1 , 𝐴2 , … . 𝐴𝑚 , 𝐵1 , 𝐵2 , … . , 𝐵𝑚 and 𝐶1 , 𝐶2 , … . 𝐶𝑚 are constants. Linear Factors of g (x): For each linear factor of g(x), the partial fraction decomposition must include the sum: 𝐴1 𝐴 𝐴𝑚 + +⋯+ 1 2 (𝑎𝑥 + 𝑏) (𝑎𝑥 + 𝑏) (𝑎𝑥 + 𝑏)𝑚 Quadratic Factors of g(x): For each quadratic factor of g(x), the partial fraction decomposition must include the sum: \frac{B_{1}x+C_{1}}{(ax^{2}+bx+c)^{1}}+\frac{B_{2}x+C_{2}}{(ax^{2}+bx+c)^{2}}+...+\frac{B_{n}x+C_{n}}{(a x^{2}+bx+c)^{n}} Step 3: With the form of the partial fraction decomposition set up, multiply both sides of the equation by the Least Common Divisor to clear fractions. Step 4: Use the equation from step 3, set up a system of linear equations by equating the constant terms and equating the coefficients of like powers of x. Step 5: Solve the system of equations from step4 and substitute the solutions to the system into the partial fraction decomposition. Calculation: Consider the provided expression, \frac{-12x-29}{2x^{2}+11x+15} Here, f(x)=-12x-29 and g(x)=2x^{2}+11x+15 Factorize g(x): g(x)=2x^{2}+11x+15 =2x^{2}+6x+5x+15 =2x(x+3)+5(x+3) =(2x+5)(x+3) So, the expression can be written as: \frac{-12x-29}{2x^{2}+11x+15}=\frac{-12x-29}{(2x+5)(x+3)} Here, in g(x), there are two factors, and both are linear. So, the expression can be decomposed as: \frac{-12x-29}{(2x+5)(x+3)}=\frac{A}{2x+5}+\frac{B}{x+3} ........(1) Here, Least Common Divisor is (2x+5)(x+3) . Multiply both sides of equation (1) by the Least Common Divisor to clear fractions: (2x+5)(x+3)[\frac{-12x-29}{(2x+5)(x+3)}]=(2x+5)(x+3)[\frac{A}{2x+5}+\frac{B}{x+3}] Simplify to obtain: -12x-29=A(x+3)+B(2x+5) -12x-29=Ax+3A+2Bx+5B -12x-29=(A+2B)x+(3A+5B) Compare the coefficients of x and constant terms, A+2B=-12 and 3A+5B=-29 So, the system of equations obtained is: A+2B=-12 .....(2) And, 3A+5B=-29 ......(3) Multiply in equation (1) with -3: -3A-6B=36 ......(4) Add the equations (3) and (4) -3A-6B=36 3A+5B=-29 \overline{B=7} Substitute —7 for B in equation (2) and simplify, A+2(-7)=-12 A-14=-12 A=2 Substitute the obtained values of A and B in equation (1): \frac{-12x-29}{(2x+5)(x+3)}=\frac{2}{2x+5}+\frac{-7}{x+3} Therefore, the partial fraction decomposition for \frac{-12x-29}{2x^{2}+11x+15} is \frac{2}{2x+5}+\frac{7}{x+3} Question #7 To calculate: The calcium present in 1 cup of milk and in 1 cup of cooked spinach if one day Shenika had 3 cups of milk and 1 cup of cooked spinach for a total of 1140 mg of calcium. The next day, she had 2 cups of milk and 1\frac{1}{2} cups of cooked spinach for a total of 960 mg of calcium. Answer: Calculation: Consider that Shenika had 3 cups of milk and 1 cup of cooked spinach for a total of 1140 mg of calcium. The next day, she had 2 cups of milk and 1\frac{1}{2} cups of cooked spinach for a total of 960 mg of calcium. Convert this problem into two system of equations. Let the amount of calcium present in 1 cup of milk be x mg and the amount of calcium present in 1 cup of cooked spinach be y mg. Shenika had 3 cups of milk and 1 cup of cooked spinach for a total of 1140 mg of calcium. Now, this can be represented by a linear equation which is as follows, 3x+y=1140 .....(1) The next day, she had 2 cups of milk and 1\frac{1}{2} cups of cooked spinach for a total of 960 mg of calcium. Now, this can be represented by a linear equation which is as follows, 2x+1.5y=960 Multiply with 10 to clear decimals of above equation. 20x+15y=9600 ......(2) 4x+3y=1920 So, the system of equation becomes as, 3x+y= 1140 4x+3y=1920 Now, multiply the first equation with -3. -9x-3y =-3420 4x+3y=1920 Add these equations and solve for x. -9x+4x-3y+3y=-3420+1920 —5x=-1500 x=\frac{-1500}{-5} x=300 Substitute 300 for x in the (1) equation and solve for y. 3(300)+y=1140 900+y=1140 y=1140-900 y=240 Substitute the ordered pair in each equation to verify the system of equation. Substitute (x,y)=(300,240) in first equation. 3\cdot300+240\overset{?}{=}1140 900+240\overset{?}{=}1140 1140=1140 The result is true. Substitute (x,y)=(300,240) in second equation. 4\cdot300+240\overset{?}{=}1920 1200+720\overset{?}{=}1920 1920=1920 The result is true. Therefore, the calcium present in 1 cup of milk is 300 mg and in 1 cup of cooked spinach is 240 mg. Question #8 To calculate: The solution for the system of equations 5x^{2}+y^{2}=14 and x^{2}-2y^{2}=-17 . If the system does not have a unique solution, determine whether the system is inconsistent, or the equations are dependent. Answer: Calculation: Consider the provided system of equations: 5x^{2}+y^{2}=14 .......(1) And, x^{2}-2y^{2}=-17 .....(2) Multiply by 2 in equation (1) 10x^{2}+2y^{2}=28......(3) Add equations (3) and (2): 11x^{2}=11 x^{2}=1 x=\pm1 Now, substitute 1 for xin equation (2) and obtain the value of y: (1)^{2}-2y^{2}=-17 1-2y^{2}=-17 -2y^{2}=-18 y^{2}=9 y=\pm3 Again, put x= and y=-3 in equation 5x^{2}+y^{2}=14 : 5(1)^{2}+(-3)^{2}\overset{?}{=}14 5+9\overset{?}{=}14 14\overset{?}{=}14 Put x=1 and y=—3 in equation x^{2}-2y^{2}=-17 (1)^{2}-2(-3)^{2}\overset{?}{=}-17 1-18\overset{?}{=}-17 -17\overset{?}{=}-17 Again, put x=-1 and y=3 in equation 5x^{2}+y^{2}=14 5(-1)^{2}+(3)^{2}\overset{?}{=}14 5+9\overset{?}{=}14 14\overset{?}{=}14 Put x=-1 and y=3 in equation x^{2}-2y^{2}=-17 (-1)^{2}-2(3)^{2}\overset{?}{=}-17 1-18\overset{?}{=}-17 -17\overset{?}{=}-17 Again, put x=-1 and y=-3 in equation 5x^{2}+y^{2}=14 5(-1)^{2}+(-3)^{2}\overset{?}{=}14 5+9\overset{?}{=}14 14\overset{?}{=}14 Put x=-1 and y=-3 in equation x^{2}-2y^{2}=-17 (-1)^{2}-2(-3)^{2}\overset{?}{=}-17 1-18\overset{?}{=}-17 -17\overset{?}{=}-17 Therefore, the solutions to the set of equations 5x^{2}+y^{2}=14 and x^{2}-2y^{2}=-17 is \{(1,3),(1,-3),(1,3),(-1,-3)\} . Question #9 To calculate: The solution set of the polynomial equation 4x^{2}-6x^{2}-20x+30=0. Answer: Formula used: According to zero product property if, mn=0, Then, m=0 Or, n=0 Calculation: This is a polynomial equation with one side equal to zero. 4x^{2}-6x^{2}-20x+30=0. Do the factors of the given equation, 2x^{2}(2x-3)-10(2x-3)=0 (2x-3)(2x^{2}-10)=0 Set each factor equal to zero. (2x-3)=0 x=\frac{3}{2} Or (2x^{2}-10)=0 x^{2}=\frac{10}{2} x=\pm\sqrt{\frac{10}{2}} x=\pm\sqrt{5} Check at x=\frac{3}{2} 4(\frac{3}{2})^{3}-6(\frac{3}{2})^{2}-20(\frac{3}{2})+30\overset{?}{=}0 \frac{54}{4}-\frac{54}{4}-30+30\overset{?}{=}9[0] 0=0 The left side value is equal to the right-side value. Thus, the value x=\frac{3}{2} is verified. Check at x=\sqrt{5} 4(\sqrt{5})^{3}-6(\sqrt{5})^{2}-20(\sqrt{5})+30\overset{?}{=}0 4\cdot5\sqrt{5}-6\cdot5-20\sqrt{5}+30\overset{?}{=}0 20\sqrt{5}-30-20\sqrt{5}+30\overset{?}{=}0 0=0 The left side value is equal to the right-side value. Thus, the value x=\sqrt{5} is verified. Check at x=-\sqrt{5} 4(-\sqrt{5})^{3}-6(-\sqrt{5})^{2}-20(-\sqrt{5})+30\overset{?}{=}0 -4\cdot5\sqrt{5}-6\cdot5+20\sqrt{5}+30\overset{?}{=}0 -20\sqrt{5}-30+20\sqrt{5}+30\overset{?}{=}0 0=0 The left side value is equal to the right-side value. Thus, the value x=-\sqrt{5} is verified. Therefore, all together the solution set of polynomial equation 4x^{3}-6x^{2}-20x+30=0 \ is \{\frac{3}{2},\pm\sqrt{5}\}. Question #10 To calculate: The Solution set of the equation -2=|x-3|-6 . Answer: Calculation: Consider the provided equation, -2=|x-3|-6 Now, add 6 to the both sides of the above equation and solve. -2=|x-3|-6 -2+6=|x-3|-6+6 4=|x-3| According to absolute value property, two cases arise: 4=x-3 or 4=-(x-3) Simplify the above expressions, x=7\ or\ x=-1 Check: Substitute the value of x=7 in the equation -2=|x-3|-6 . -2\overset{?}{=}|7-3|-6 -2\overset{?}{=}|4|-6 -2\overset{?}{=}4-6 -2\overset{?}{=}-2 which is true. Now, substitute the value of x=-1 in the equation -2= |x-3|-6 -2\overset{?}{=}|-1-3|-6 -2\overset{?}{=}|4|-6 -2\overset{?}{=}4-6 -2\overset{?}{=}-2 which is true. Hence, the solution set of the equation -2=|x-3|-6 is (-1.7). Question #11 To calculate: The solution of the equation 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0. Answer: Calculation: Consider the equation, 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0. Now, simplify the equation 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0. 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0. 2\{(\frac{x}{3})^{2}+(1)^{2}+2(\frac{x}{3})(1)\}+5(\frac{x}{3}+1)-12=0. 2\{(\frac{x^{2}}{9})+1+\frac{2x}{3}\}+\frac{5x}{3}+5-12=0. \frac{2x^{2}}{9}+\frac{9x}{3}-5=0 Now, taking the least common multiple of the denominator, \frac{2x^{2}}{9}+\frac{9x}{3}-5=0 \frac{2x^{2}+27x-45}{9}=0 2x^{2}+27x-45=0 This is a quadratic equation so compare the equation from the standard form of quadratic equation ax^{2}+bx-c=0 and identify the values of a, b and c. Here, a=2, b=27 and c=-45 Now, apply the quadratic formula, x=\frac{-(27)\pm\sqrt{(27)^{2}-4(2)(-45)}}{2(2)} =\frac{-27\pm\sqrt{729+360}}{4} =\frac{-27\pm\sqrt{1089}}{4} =\frac{-27\pm33}{4} Firstly, consider the positive sign, x=\frac{-27+33}{4} =\frac{6}{4} =\frac{3}{2} Now, consider the negative sign, x=\frac{-27-33}{4} =\frac{-60}{4} =-15 Now, to check the solution put the values of x in the original equation, Substitute x=-15 in the equation 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0 2(\frac{-15}{3}+1)^{2}+5(\frac{-15}{3}+1)-12\overset{?}{=}0 2(-5+1)^{2}+5(-5+1)-12\overset{?}{=}0 2(-4)^{2}+5(-4)-12\overset{?}{=}0 Further simplified, 2(16)-20-12\overset{?}{=}0 32-32\overset{?}{=}0 0=0 This solution is correct. Now, substitute x=\frac{3}{2} in the equation 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0 2(\frac{\frac{3}{2}}{3}+1)^{2}+5(\frac{\frac{3}{2}}{3}+1)-12\overset{?}{=}0 2(\frac{1}{2}+1)^{2}+5(\frac{1}{2}+1)-12\overset{?}{=}0 2(\frac{3}{2})^{2}+5(\frac{3}{2})-12\overset{?}{=}0 Further simplified, 2(\frac{9}{4})+5(\frac{3}{2})-12\overset{?}{=}0 \frac{9}{2}+\frac{15}{2}-12\overset{?}{=}0 \frac{24-24}{2}\overset{?}{=}0 0=0 This solution is correct. Therefore, the solution set of the equation 2(\frac{x}{3}+1)^{2}+5(\frac{x}{3}+1)-12=0 is \{-15, \ \frac{3}{2}\}. Question #12 To calculate: The solution for the system of provided equations: 5a - 2b + 3c = 10 -3a + b - 2c = -7 a + 4b - 4c = -3 Answer: (A) 5a - 2b +3c = 10 (B) -3a + b - 2c = -7 (C) a + 4b - 4c -3 Consider (A) or (B): (A) 5a - 2b + 3c = 10 (B) -3a + b - 2c = -7 Multiply (B) with 2: (A) 5a - 2b + 3c = 10 2 \cdotask (B) - 6a + 2b - 4c = -14 Now, add (A) and (B): (D) - a - c = -4 Consider (A) and (C): (A) 5a - 2b + 3c = 10 (C) a + 4b - 4c = -3 Multiply (A) with 2: 2 \cdotask (A) 10a - 4b + 6c = 20 (C) a + 4b - 4c = -3 Add both the above equastion: (E) 11a + 2c = 17 Now consider (D) and (E): (D) - a - c = -4 (E) 11a + 2c = 17 Multipy (D) with 2: 2 \cdotask (D) - 2a - 2c = -8 (E) 11a + 2c = 17 Now, add both the equastions and solve for a: 9a = 9 (dividing both sides by 9) a=1 Substitute 1 for a in (D): -(1) -c = -4 Adding 1 to both sides and then dividing by -1, -c = -3 c=3 Substitude 3 for c and 1 for a in (A) and solve for b: 5 (1) - 2b + 3 (3) = 10 -2b + 14 = 10 Substracting 14 from both sides and then dividing by -2, -2b = -4 b=2 -2b=-4 b=2 So, the values obtained are a=1, b=2 and c=3. Substitute thease values in each of the provided equations to verify: First Equation: 5(1) - 2(2) + 3(3) 10 5 - 4 + 9 10 10 10 The result is true. Second Equation: -3 (1) + (2) - 2(3) -7 -3 + 2 -6 -7 -7 -7 The result is true. Third Equation: (1) + 4 (2) - 4(3) -3 1 + 8 - 12 -3 -3 -3 The result is true. 5a - 2b + 3c = 10 Therefore, the solution of the system of equations -3a + b - 2c = -7 is a + 4b - 4c = -3 Question #13 To calculate: The pardital decomposition of the function \frac{-12x-29}{2x^{2}+11x+15} Answer: Given Information: The provided expression is: \frac{\text{-12}x-29}{\text{2x^+11x}+15} . Formula used: Decomposition of \frac{f(x)}{g(x)} into Partial Fractions: Consider a rational expression \frac{f(x)}{g(x)} , where f(x) and g(x) are polynomial with real coefficients, \cancel{c} (x) \ne 0 , and the degree of f(x) is less than degree of g(x). Step 1: Factor the denominator g (x) completely into linear factors of the form \(ax + b)^{m} and quadratic factors of the form ( ax^{2} + bx +c)^{n} that are not further factorable over the integers. Step 2: Set up the form of decomposition. That is, write the original rational expression \frac{f(x)}{g(x)} as a sum of simpler fractions using these guidelines. Note that A_{1} ,A_{2},.....,A_{m},B_{1},B_{2},.....,B_{m} and C_{1},C_{2},....,C_{m} are constants. Linear Factors of g(x): For each linear factor of g(x), the partial fraction decomposition must include the sum: \frac{A_1}{(ax+b)^{1}}+\frac{A_2}{(ax+b)^{2}}+...+\frac{A_m}{(ax+b)^{m}} Quadratic Factors of g (x): For each quadratic factor of g (x), the partial fraction decomposition must include the sum: \frac{B_{1}x+C_{1}}{(ax^{2}+bx+c)^{1}}+\frac{B_{1}x+C_{1}}{(ax^{2}+bx+c)^{1}}+...+\frac{B_{1}x+C_{1}}{(a x^{2}+bx+c)^{1}} Step 3: With the form of the partial fraction decomposition set up, multiply both sides of the equation by the Least Common Divisor to clear fractions. Step 4: Use the equation from step3, set up a system of linear equations by equating the constant terms and equating the coefficients of like powers ofx. Step 5: Solve the system of equations from step4 and substitute the solutions to the system into the partial fraction decomposition. Calculation: Consider the provided expression, \frac{-12x-29}{2x^{2}+11x+15} Here, f (x) =-12x—29 and g(x) = 2x^{2} + 11x + 15 Factorize g(x): g(x)=2x^{2}+11x+15 =2x^{2}+6x+5x+15 =2x(x+3)+5(x+3) =(2x+5)(x+3) So, the expression can be written as: \frac{-12x-29}{2x^{2}+11x+15}=\frac{-12x-29}{(2x+5)(x+3)} Here, in g(x), there are two factors, and both are linear. So, the expression can be decomposed as: \frac{-12x-29}{(2x+5)(x+3)}=\frac{A}{2x+5}+\frac{B}{x+3} ............(1) Here, Least Common Divisor is (2x + 5) (x + 3) . Multiply both sides of equation (1) by the Least Common Divisor to clear fractions: \[(2x+5)(x+3)\begin{bmatrix}-12x-29 \\(2x+5) (x+3) \end{bmatrix}=(2x+5)(x+3) [\frac{A}{2x+5}+\frac{B}{x+3}]\] Simplify to obtain: -12x- 29 = A(x+3)+B(2z+5) -12x- 29 =Ax+3A + 2Bx + 5B -12x-29 = (A + 2B) x + (3A + 5B) Compare the coefficients of x and constant terms, A+2B = -12 and 3A + 5B = -29 So, the system of equations obtained is: A+2B = -12 ............(2) And, 3A +5B = 29 ............ (3) Multiply in equation (1) with -3: -3A - 6B = 36 .......(4) Add the equations (3) and (4) -3A - 6B = 36 frac{3A+5B=-29}{B=7}\ Substitute -7 for B in equation (2) and simplify, A+2(-7) =-12 A-14=-12 A=2 Substitute the obtained values of A and B in equation (1): \\frac{-12x-29}{(2x+5)(x+3)}=\frac{2}{2x+5}+\frac{-7}{x+3} Therefore, the partial fraction decomposition for \\frac{-12x-29}{2x^{2}+11x+15}\ is\ \frac{2}{2x+5}+\frac{7}{x+5} . Question #14 To calculate: The partial decomposition of \frac{2x^{3}-x^{2}+8x-16}{x^{4}+5x^{2}+4} . Answer: Formula used: Decomposition of \frac{f(x)}{g(x)} Partial Fractions: Consider a rational expression \frac{f(x)}{g(x)} , where f(x) and g (x) are polynomial with real coefficients, g (x)\ne 0, and the degree of f (x) is less than degree of g(x). Step1: Factor the denominator g (x) completely into linear factors of the form \(ax + b)^{m} and quadratic factors of the form \(ax^{2}+ bx +c)^{n} that are not further factorable over the integers. ‘Step2: Set up the form of decomposition. That is, write the original rational expression \frac{f(x)}{g(x)} as a sum of simpler fractions using these guidelines. Note that A1,A2,.....,Am,B1,B2,....,Bm and C1,C2,.....,Cm are constants. Linear Factors of g(x): For each linear factor of g(x), the partial fraction decomposition must include the sum: \frac{A1}{(ax+b)^{1}}+\frac{A2}{(ax+b)^{2}}+....+\frac{Am}{(ax+b)^{m} Quadratic Factors of g (x): For each quadratic factor of g (x), the partial fraction decomposition must include the sum: \frac{B_{1}x+C_{1}}{(ax^{2}+bx+c)^{1}}+\frac{B_{2}x+C_{2}}{(ax^{2}+bx+c)^{2}}+...+\frac{B_{n}x+C_{n}}{(a x^{2}+bx+c)^{n}} Step3: With the form of the partial fraction decomposition set up, multiply both sides of the equation by the LCD to clear fractions. ‘Step4: Use the equation from step3, set up a system of linear equations by equating the constant terms and equating the coefficients of like powers of x. Step 5: Solve the system of equations from step4 and substitute the solutions to the system into the partial fraction decomposition. Calculation: Consider the provided expression: NCK \frac{2x^{3}-x^{2}+8x-16}{x^{4}+5x^{2}+4} Here, \f (x) = 2x^{3} - x^{2} + 8x— l6 and g(x) = x^{4} +5x^{2} +4 . Factorize g(x): \g(x)=x^{4}+5x^{2}+4 =x^{4}+x^{2}+4x^{2}+4 =x^{2}(x^{2}+1)+4(x^{2}+1) =(x^{2}+4) (x^{2}+1) So, the expression becomes: \frac{2x^{3}-x^{2}+8x-16}{(x^{2}+4)(x^{2}+1)} Here, in g (x), there are two factors, both are quadratic. So, the expression can be decomposed as: \frac{2x^{3}-x^{2}+8x-16}{(x^{2}+4)(x^{2}+1)} =\frac{Ax+B}{x^{2}+4}+\frac{Cx+D}{x^{2}+1} Here, LCD = \(x^{2} +4) (x^{2} +1) ‘Multiply both sides by the LCD to clear fractions: \(x^{2}+4)(x^{2}+1)(\frac{2x^{3}-x^{2}+8x16}{(x^{2}+4)(x^{2}+1)})=(x^{2}+4)(x^{2}+1)(\frac{Ax+B}{x^{2}+4}+\frac{Cx+D}{x^{2}+1}) Simplify to obtain: \2x^{3} - x^{2} + 8x - 16 = (Ax +B) (x^{2} +1) + (Cx+ D) (x^{2} +4) \2x^{2} - x^{2}+ 8x- 16 = Ax^{3} + Ax + Bx^{2} +B + Cx^{3} +4Cx + Dx^{2 }+ 4D \2x^{2} - x^{2}+ 8x- 16 = (A+ C)x^{3} + (B+ D)x^{2} + (A +4C) x + (B+ 4D) Compare the coefficients of variable terms and constant terms: \A+C=2, B+ D = -1,A+4C =8 and B+4D =-16 Consider A+ C =2 and A+4C =8: Subtract A + 4C = 8 from A +C = 2: -3C=-6(C=2)(Divide both sides by -3) Substitute 2 for C in A+C=2: A+2=2(A=0) (Subtract 2 from both sides) Consider B + D =-1 and B+4D =-16: Subtract B + D = -I from B+ 4D = -16: 3D =I5(D=-5) (Divide both sides by 3) Substitute -5 for D in B+ D =-1: B-5=-1(B=4) (Add 5 on both sides) Substitute the values of A, B, C and D in equation(1): \frac{2x^{3}-x^{2}+8x-16}{(x^{2}+4)(x^{2}+1)}=\frac{4}{x^{2}+4}+\frac{2x-5}{x^{2}+1} Therefore, the partial fraction decomposition for \frac{2x^{3}-x^{2}+8x-16}{x^{2}+5x^{2}+4}\ is\ \frac{4}{x^{2}+4}+\frac{2x-5}{x^{2}+1} . Question #15 To calculate: Thesolution set of the equation \sqrt{2d}= 1 -\sqrt{d +7} Answer: Consider the given equation, \sqrt{2d}= 1 -\sqrt{d +7} Now, apply Square on the both sides of the above equation, \sqrt{2d}= 1 -\sqrt{d +7} (\sqrt{2d})^2= (1 -\sqrt{d +7})^2\ 2d = 1-2\sqrt{d+7}+d+7 2d = 8 + d -2\sqrt{d+7} Simplify the above equation, 2d = 8 + d -2\sqrt{d+7} 2d-d=8-2\sqrt{d+7} 2\sqrt{d+7}= 8-d Apply square on both sides: of the above equation, (2\sqrt{d+7})^2=(8-d)^2 4(d+7)=64-16d+d^2 4d+28=64-16d+d^2 Now subtract 4d+28 from both sides of the equation, 4d+28=64-16d+d^2 4d+28-(4d+28)=64-16d+d^2-(4d+28) 0=64-16d+d^2-4d-28 0=d^2-20d+36 Compare the equation d^2-20d+36=0 with the general form quadratic equation ax^2+bx+c=0 . Then, a=1,b=-20,c=36 . Put the values of a,b,c in the expression, \frac{-b\pm\sqrt{b^2-4ac}}{2a} and solve, d=\frac{-(-20)\pm\sqrt{(-20)^2-4\times1\times36}}{2\times1} =\frac{20\pm\sqrt{400-144}}{2} =\frac{20\pm\sqrt{256}}{2} =\frac{20\pm16}{2} Therefore, d=10\pm8 . That is, d=18 or d=2 . Check: Put the value d=18 in the provided equation \sqrt{2d}=1-\sqrt{d+7} \sqrt{2 \times18}\overset{?}{=}1-\sqrt{18+7} \sqrt{36}\overset{?}{=}1-\sqrt{25} 6\overset{?}{=}1-5 6\overset{?}{=}-5 which is not true. Therefore, d=18 is not a solution of the equation \sqrt{2d}=1-\sqrt{d+7} . Now, put the value d=2 in the provided equation \sqrt{2d}=1-\sqrt{d+7} . \sqrt{2\times2 }\overset{?}{=}1-\sqrt{2+7} \sqrt{4}\overset{?}{=}1-\sqrt{9} 2\overset{?}{=}1-3 2\overset{?}{=}-2 which is not true. Therefore, d=2 is not a solution of the equation \sqrt{2d}=1-\sqrt{d+7} . Hence, the solution set of the equation \sqrt{2d}=1-\sqrt{d+7} is {}. Question #16 The field mouse population in Example 1 satisfies the differential equation dp/dt=0.5p-450 . (a) Find the time at which the population becomes extinct ifp(0)=850 . (b) Find the time of extinction if p(0)=p0 , where 0 0.5 . Answer: Given: H_{0}:p=0.50 H_{a}:p>0.50 z=1.52 The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, when the null hypothesis is true. Determine the P-value using the normal probability table in the appendix. P=P(Z>1.52)=1-P(Z<1.52)=1-0.9357=0.9357 Note: Command for a Ti83/84-calculator is normal cdf(1.52, 1E99) or 1E99 replaced by some other very large value. Question #19 Find the product of this question: 2x^4(3x^2 + 4x + 1) Answer: The given expression is 2x^4(3x^2 + 4x + 1) Compute the product of the given expression as follows. 2x^4(3x^2 + 4x + 1) = (2x^4 \cdot 3x^2) + (2x^4 \cdot 4x) + (2x^4 \cdot 1) [∵ By FOIL method] = 6x^{4+2} + 8x^{4+1} + 2x^4 [\bucause x^a \cdot x^b = x^{a+b}] = 6x^6 + 8x^5 + 2x^4 Thus, the product of the given expression is 6x^6 + 8x^5 + 2x^4 . Question #20 If xy+6e^{y}=6e find the value of y'' at the point where x=0 Answer: Step 1 Ans. xy+6e^{y}=6e x\frac{dy}{dx}+y+6e^{y}+\frac{dy}{dx}=0 x\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}+6e^{y}\frac{d^{2}y}{dx^{2}}+6e^{y}(\frac{dy}{dx})^{2}=0 at point x=0 0\frac{d^{2}y}{dx^{2}}+2(\frac{dy}{dx})+6e^{y}\frac{d^{2}y}{dx^{2}}+6e^{y}(\frac{dy}{dx})^{2}=0 Step 2 6e^{y} y'' =-2y'-6e^{y}(y')^{2} y''=\frac{-2y'}{6e^{y}}-\frac{6e^{y}(y')^{2}}{6e^{y}} \[y''=-\frac{1}{3}(\begin{array}{c}y'\\ e^{y}\end{array})-(y')^{2}\]