Questions and Answers #21 Equations

Questions and Answers Sheet 21 Equations Question #1 How do you solve for x in the equation 𝑥𝑦 = 𝑥 + 𝑦 Answer: xy=x+y -x-x x(y-1)=y 𝑦 𝑥= 𝑦−1 Question #2 How do you solve |−10𝑥 − 5| = 0 Answer: The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent. However, because -0 equals 0 we can just solve the term within the absolute value function once for 0: -10x-5=0 -10x-5+5=0+5 -10x-0=5 −10𝑥 −5 = −10 −10 −10𝑥 1 =− −10 𝑥=− 1 2 2 Question #3 How do you solve for x in 𝑥 = 18 𝑥+3 Answer: Explanation: 18 𝑥= 𝑥+3 ⇒ 𝑥(𝑥 + 3) = 18 ⇒ 𝑥 2 + 3𝑥 − 18 = 0 ⇒ 𝑥 2 + 6𝑥 − 3𝑥 − 18 = 0, [6 × 3 = 18, 6 − 3 = 3] ⇒ 𝑥(𝑥 + 6) − 3(𝑥 + 6) = 0 ⇒ (𝑥 + 6)(𝑥 − 3) = 0 ⇒ 𝑥 = −6, 𝑥 = 3 Clearly, these satisfy the given eqn. hence, the roots are confirmed. Question #4 How do you solve for y: 2𝑥 − 3𝑦 = 6 ? Answer: Explanation: You must do the same action on both sides of the = sign; 2x=6+3y 2x-6=3y 2𝑥−6 =𝑦 3 Therefore: 2 𝑦 = ( )𝑥 − 2 3 Question #5 How do you solve 2𝑥 = 0.5 ? Answer: We know that for any 𝑎 𝑥 = 𝑏, log 𝑎 𝑏 = 𝑥 Here, a=2 and b=0.5 So we take log 2 0. 5 = −1 So x=-1 Question #6 How do you factor the trinomial 2𝑥 2 + 20𝑥 + 50 ? Answer: Explanation: We see that 2 is a common factor among all of the coefficients so we can factor 2 to the front giving us 2(𝑥 2 + 10𝑥 + 25) We now need 2 numbers which add to give 10 and multiply to give 25 (or use the quadratic formula). These 2 numbers will be 5 and 5(5 + 5 = 10,5 ⋅ 5 = 25) . So we get 2(𝑥 + 5)(𝑥 + 5) = 2(𝑥 + 5)2 Question #7 How do you solve the system of equations x-y=7 and -2x+5y=-8 ? Answer: Your goal here is to remove one of the variables from the problem. You can see that the first equation has x and the second equation has -2x. If we double the first equation, we get: 2x-2y=14 Then we simply add that to the second equation: 2x-2y=14 ±2𝑥 + 5𝑦 = −8 3y=6 The positive 2x and the negative 2x cancel out, leaving us with just the 3y = 6 . Divide both sides by 3 and we get y=2 Last, just plug 2 in for y in either equation (I'll choose the first since it's simpler): x-2=7 Add 2 to both sides to get x=9 So our final answer is x=9,y=2 Question #8 How do I find the binomial expansion of (2𝑥 + 1)3 ? Answer: We must use our knowledge of the binomial expansion: Method 1: We can use: 𝑛(𝑛 − 1) 2 𝑛(𝑛 − 1)(𝑛 − 2) 3 (𝑥 + 1)𝑛 = 1 + 𝑛𝑥 + 𝑥 + 𝑥 +⋯ 2! 3! Substituting n=3 and 𝑥 𝑓𝑜𝑟 2𝑥 ⇒ 3⋅2 3⋅2⋅1 (2𝑥 + 1)3 = 1 + (3 ⋅ 2𝑥) + ⋅ (2𝑥)2 + ⋅ (2𝑥)3 2! 3! = 1 + 6𝑥 + 12𝑥 2 + 8𝑥 3 Question #9 1 1 2 4 How do you solve 𝑥 − = 1 ? Answer: Solve: 1 1 𝑥− =1 2 4 1 Convert 1 to an improper fraction by multiplying the denominator by the whole number 1, and adding 4 the numerator 1. Set the result over the denominator 4. 4×1+1 4 = 5 4 Rewrite the equation. 1 5 𝑥− = 2 Add 1 4 to both sides. 2 5 1 𝑥= + 4 2 In order to add or subtract fractions, the denominators must be the same. In this case the least common 1 denominator (LCD) is 4. Multiply by a fraction equal to 1 in order to produce an equivalent fraction. For example 5 5 1 2 = 1 . This way the value of the fraction is not changed. Multiply by 2 5 1 2 𝑥= + × 4 2 2 Simplify. 7 𝑥= 2 2 4 Question #10 1 How do you solve 2𝑥(𝑥 − 5)−1 + = 0 𝑥 Answer: Explanation: 1 2(𝑥 − 5)−1 + = 0 𝑥 Re-arrange the terms to have one fraction on each side. (note that the negative index moves the bracket to the denominator) 𝑥 ≠ +5 and 𝑥 ≠ 0 2 −1 = 𝑥−5 𝑥 2𝑥 = −𝑥 + 5 3𝑥 = 5 5 𝑥= 3 Question #11 How do you solve the simultaneous equations m+n=19 and m-n=7 Answer: Suppose the numbers are m and n We are given: (𝑚 + 𝑛 = 19), (𝑚 − 𝑛 = 7): Adding these two equations together we get: 2m=26 Dividing both sides by 2, we find m=13 Then from the first equation, we find: n=19-m=6 Question #12 How do you solve 9𝑒 + 4 = −5𝑒 + 14 + 13𝑒 Answer: First, group and combine like terms on the right side of the equation: 9𝑒 + 4 = −5𝑒 + 13𝑒 + 14 9𝑒 + 4 = (−5 + 13)𝑒 + 14 9𝑒 + 4 = 8𝑒 + 14 Now, subtract 4 and 8e from each side of the equation to solve for e while keeping the equation balanced: 9𝑒 − 8𝑒 + 4 − 4 = 8𝑒 − 8𝑒 + 14 − 4 (9 − 8)𝑒 + 0 = 0 + 10 1𝑒 = 10 𝑒 = 10 Question #13 How do you write 𝑦 = 𝑥 2 + 16𝑥 + 14in vertex form? Answer: By completing the square: 16 2 16 2 2 2 𝑦 = 𝑥 2 + 16𝑥 + ( ) − ( ) + 14 𝑦 = (𝑥 + 8)2 − 64 + 14 𝑦 = (𝑥 + 8)2 − 50 Question #14 What are the roots of the equation 𝑥 2 − 5𝑥 + 1 = 0 ? Answer: You can solve this equation using 2 methods, one being completing the square method, and the other by using the quadratic formula. 1) Commencing completing the square method, 𝑥 2 − 5𝑥 + 1 = 0 Subtract 1 on both sides, 𝑥 2 − 5𝑥 = −1 Add 6.25 on both sides, 𝑥 2 − 5𝑥 + 6.25 = −1 + 6.25 Apply perfect quadratic square formula, (𝑥 − 2.5)2 = 5.25 Square root both sides, 𝑥 − 2.5 = ±√5.25 Add 2.5 to both sides, 𝑥 = ±√5.25 + 2.5 Hence, x=4.79128784 or 0.20871215 2) Commencing quadratic formula method, 𝑥 2 − 5𝑥 + 1 = 0 Quadratic equation, 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 Substitute a=1,b=-5, c=1 into the quadratic formula. 𝑥= 5±√21 2 Hence, x=4.79128784 or 0.20871215 Question #15 How do you solve |𝑥 + 4| = 6 ? Answer: The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent. Solution: x+4=-6 x+4-4=-6-4 x+0=-10 x=-10 Solution 2 x+4=6 x+4-4=6-4 x+0=2 x=2 The solutions Are: x=-10 and x=2 Question #16 How does one verify cos 2 x − sin 2 x = cos 2 x ? 2 1 − tan x cos 2 x − sin 2 x = cos 2 x 2 1 − tan x I know that there are a few properties in here, but I don't know how the signs would effect them. Answer: cos 2 𝑥 − sin2 𝑥 1 − tan2 𝑥 cos2 𝑥−sin2 𝑥 = sin2 𝑥 1− cos2 𝑥 (cos 2 𝑥 − sin2 𝑥) = cos 2 𝑥 − sin2 𝑥 cos 2 𝑥 = cos 2 𝑥 Proved. Question #17 Solve the equation 3𝑝 4 =6? Answer: Explanation: 3𝑝 =6 4 Multiplying by 4 on both sides, we get 3𝑝 ⋅4=6⋅4 4 3p=24 Dividing by 3 to isolate p, we get 3𝑝 24 = 3 3 p=8 Question #18 2 What is the equivalent to ? 5 Answer: Explanation: 𝑎 can be written as 𝑏 2 5 = Or 4 = 10 2𝑛 80 200 𝑎𝑛 𝑏𝑛 so =⋯ 5𝑛 Question #19 How do you solve for x in ax-c=b ? Answer: First, add c to each side of the equation to isolate the x term while keeping the equation balanced: ax-c+c=b+c ax-0=b+c ax=b+c Now, divide each side of the equation by a to solve for x while keeping the equation balanced: 𝑎𝑥 𝑎 = 𝑏+𝑐 𝑎 𝑎𝑥 𝑏 + 𝑐 = 𝑎 𝑎 𝑏+𝑐 𝑥= 𝑎 Or 𝑏 𝑐 𝑎 𝑎 𝑥= + Question #20 Which is 𝑛 ⇒ 𝑒𝑟? 𝑓(𝑥) = 2𝑥 2 + 3𝑥 or 𝑔(𝑥) = 𝑥 2 + 4 Answer: Let us write these equations of parabolas in their vertex form i.e. 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 where (h.k) is the vertex and a is quadratic coefficient. The greater the quadratic coefficient, the 𝑛 ⇒ 𝑒r is the parabola. 3 𝑓(𝑥) = 2𝑥 2 + 3𝑥 = 2 (𝑥 2 + 𝑥) 2 3 3 2 3 2 = 2 (𝑥 2 + 2 × 𝑥 + ( ) ) − 2 × ( ) 4 4 4 3 2 9 = 2 (𝑥 + ) − 4 8 and 𝑔(𝑥) = 𝑥 2 + 4 = (𝑥 − 0)2 + 4 To find whether a parabola is 𝑛 ⇒ or wide, we should look at the quadratic coefficient of the parabola, which is 2 in f(x) and 1 in g(x) and hence 𝑓(𝑥) = 2𝑥 2 + 3𝑥 is 𝑛 ⇒ 𝑒𝑟 Question #21 Solve for 𝑥: 16 ⋅ 4𝑥+2 − 16 ⋅ 2𝑥+1 + 1 = 0 ? Answer: 16 ⋅ 4𝑥+2 − 16 ⋅ 2𝑥+1 + 1 = = 162 ⋅ 4𝑥 − 2 ⋅ 16 ⋅ 2𝑥 + 1 = 162 (2𝑥 )2 − 32 ⋅ 2𝑥 + 1 = 0 Making 𝑦 = 2𝑥 162 𝑦 2 − 32𝑦 + 1 = 0 solving for y we have 1 𝑦 = = 2𝑥 = 2−4 then x=-4 16