Linear Equation1.Word problemsThe difficult part of solving word problems is translating the words into equations, it's good to work with lots of different problems. I'll work through a variety of problems below.Example.68 less than 5 times a number is equal to the number. Find the number.Let x be the number. Note that "68 less than 5 times the number" translates to the expression,not. So the problem statement gives

Example.Calvin Butterball buys a book for $14.70, which is adiscount off the regular price. What is the regular price of the book?Let x be the regular price. Adiscount is, so the discounted price is. Set this equal to 14.7 and solve for x:

2.digit problemsIf the problem involves interchanging of the digits in the integer then you would need to convert from the digits to numbers and vice versa. To convert the digits to numbers, we need to multiply with the digit with theplace value of the digit.For example, the value of the number formed by the digit 4 in the tens place and the digit 3 in the ones place is 4 10 + 3 1Example:The sum of the digits of a two-digit number is 11. If we interchange the digits then the new number formed is 45 less than the original. Find the original number.Solution:Step 1:Assign variables :Letx=ones digit

t=tens digit

Sentence: The sum of the digits of a two-digit number is 11.x+t =11Isolatevariablexx= 11 t (equation 1)Step 2:Convert digits to numberOriginal number =t 10 +xInterchanged number =x 10 +tSentence: If we interchange the digits then the new number formed is 45 less than the original.Interchanged = Original 45x 10 +t = t 10 +x 4510x+t =10t+x 4510xx+t= 10t 45(x to both sides)10xx= 10tt 45( t to both sides)10xx+ 45 = 10tt(+ 45 to both sides)10tt= 10xx+ 45(Rewrite equation with t on the left hand side)Combine like terms10tt =10xx+ 459t= 9x+ 45 (equation 2)Substituteequation 1intoequation 29t= 9(11 t) + 459t= 99 9t+ 45Isolatevariablet9t+ 9t= 99 + 4518t= 144

The tens digit is 8. The ones digit is 11 8 = 3Answer:The number is 83.

3.Investment problemsInvestment problems usually involve simple annual interest (as opposed tocompounded interest), using the interest formulaI = Prt, whereIstands for the interest on the original investment,Pstands for the amount of the original investment (called the "principal"),ris the interest rate (expressed in decimal form), andtis the time.You put$1000into an investment yielding6%annual interest; you left the money in for two years. How much interest do you get at the end of those two years?In this case,P= $1000,r= 0.06(because I have to convert the percent to decimal form), and the time ist= 2.Substituting, I get:I = (1000)(0.06)(2) = 120I will get$120in interest.

examplesYou put$1000into an investment yielding6%annual interest; you left the money in for two years. How much interest do you get at the end of those two years?In this case,P= $1000,r= 0.06(because I have to convert the percent to decimal form), and the time ist= 2.Substituting, I get:I = (1000)(0.06)(2) = 120I will get$120in interestYou invested$500and received$650after three years. What had been the interest rate?For this exercise, I first need to find the amount of the interest. Since interest is added to the principal, and sinceP= $500,thenI= $650 500 = $150. The time ist= 3. Substituting all of these values into the simple-interest formula, I get:150 = (500)(r)(3)150 = 1500r150/1500=r= 0.104.Mixture problemsMixture problems involve creating a mixture from two or more things, and then determining some quantity (percentage, price, etc) of the resulting mixture.Example: Your school is holding a "family friendly" event this weekend. Students have been pre-selling tickets to the event; adult tickets are$5.00,and child tickets (for kids six years old and under) are$2.50. From past experience, you expect about13,000people to attend the event. But this is the first year in which tickets prices have been reduced for the younger children, so you really don't know how many child tickets and how many adult tickets you can expect to sell. Your boss wants you to estimate the expected ticket revenue. You decide to use the information from the pre-sold tickets to estimate the ratio of adults to children, and figure the expected revenue from this information.You consult with your student ticket-sellers, and discover that they have not been keeping track of how many child tickets they have sold. The tickets are identical, until the ticket-seller punches a hole in the ticket, indicating that it is a child ticket. But they don't remember how many holes they've punched. They only know that they've sold548tickets for$2460. How much revenue from each of child and adult tickets can you expect?LetAstand for the number of adult tickets pre-sold, andCstand for the child tickets pre-sold. ThenA+C= 548. Also, since each adult ticket cost$5.00, then($5.00)Astands for the revenue brought in from the adult tickets pre-sold; likewise,($2.50)Cstands for the revenue brought in from the child tickets. Then the total income so far is given by($5.00)A+ ($2.50)C= $2460. But we can only solve an equation with one variable, not two. So look again at that first equation. IfA+C= 548, thenA= 548 C(orC= 548 A; it doesn't matter which variable you solve for). Organizing this information in a grid, we get:tickets sold$/tickettotal $

adult548 C$5$5(548 C)

childC$2.50$2.50C

total548---$2460

From the last column, we get (total $ from the adult tickets) plus (total $ from the child tickets) is (the total $ so far), or, as an equation:($5.00)(548 C) + ($2.50)C= $2460$2740 ($5.00)C+ ($2.50)C= $2460$2740 ($2.50)C= $2460($2.50)C= $280C=$280/$2.50= 112Then112child tickets were pre-sold, soA= 548 112 = 436adult tickets were sold. (Using "A" and "C" for our variables, instead of "x" and "y", was helpful, because the variables suggested what they stood for. We knew instantly that "C= 112" meant "112child tickets". This is a useful technique.)Now we need to figure out how many adult and child tickets we can expect to sell overall. Since436out of548pre-sold tickets were adult tickets, then we can expect436/548, or about79.6%, of the total tickets sold to be adult tickets. Since we expect about13,000people, this works out to about10,343adult tickets. (You can find this value by usingproportions, by the way.) The remaining2657tickets will be child tickets. Then the expected total ticket revenue totals to$58,357.50, of which($5.00)(10,343) =$51,715will come from adult tickets, and($2.50)(2,657) =$6,642.50will come from child tickets.Example: Another one. This time, suppose you work in a lab. You need a15%acid solution for a certain test, but your supplier only ships a10%solution and a30%solution. Rather than pay the hefty surcharge to have the supplier make a15%solution, you decide to mix10%solution with30%solution, to make your own15%solution. You need10liters of the15%acid solution. How many liters of10%solution and30%solution should you use?Letxstand for the number of liters of10%solution, and letystand for the number of liters of30%solution. (The labeling of variables is, in this case, very important, because "x" and "y" are not at all suggestive of what they stand for. If we don't label, we won't be able to interpret our answer in the end.) For mixture problems, it is often very helpful to do a grid:

liters sol'npercent acidtotal liters acid

10%sol'nx0.100.10x

30%sol'ny0.300.30y

mixturex+y= 100.15(0.15)(10) = 1.5

Sincex+y= 10, thenx= 10 y. Using this, we can substitute forxin our grid, and eliminate one of the variables:Copyright Elizabeth Stapel 1999-2011 All Rights Reservedliters sol'npercent acidliters acid

10%sol'n10 y0.100.10(10 y)

30%sol'ny0.300.30y

mixturex+y= 100.15(0.15)(10) = 1.5

When the problem is set up like this, you can usually use the last column to write your equation: The liters of acid from the10%solution, plus the liters of acid in the30%solution, add up to the liters of acid in the15%solution. Then:0.10(10y) + 0.30y= 1.510.10y+ 0.30y= 1.51 + 0.20y= 1.50.20y= 0.5y=0.5/0.20= 2.5Then we need2.5liters of the30%solution, andx= 10 y= 10 2.5 =7.5liters of the10%solution. (If you think about it, this makes sense. Fifteen percent is closer to10%than to30%, so we ought to need more10%solution in our mix.)5. Age problems In three more years, Miguel's grandfather will be six times as old as Miguel was last year. When Miguel's present age is added to his grandfather's present age, the total is68. How old is each one now? This exercise refers not only to their present ages, but also to both their ages last year and their ages in three years, so labelling will be very important. I will label Miguel's present age as "m" and his grandfather's present age as "g". Thenm+g= 68. Miguel's age "last year" wasm 1. His grandfather's age "in three more years" will beg+ 3. The grandfather's "age three years from now" is six times Miguel's "age last year" or, in math:g+ 3 = 6(m 1)This gives me two equations with two variables:m+g= 68g+ 3 = 6(m 1)Solving the first equation, I getm= 68 g. (Note: It's okay to solve for "g= 68 m", too. The problem will work out a bit differently in the middle, but the answer will be the same at the end.) I'll plug "68 g" into the second equation in place of "m":g+ 3 = 6m 6g+ 3 = 6(68 g) 6g+ 3 = 408 6g 6g+ 3 = 402 6gg+ 6g= 402 37g= 399g= 57Since "g" stands for the grandfather's current age, thenthe grandfather is 57 years old. Sincem+g= 68, thenm= 11, andMiguel is presently eleven years old. One-half of Heather's age two years from now plus one-third of her age three years ago is twenty years. How old is she now?This problem refers to Heather's age two years in the future and three years in the past. So I'll pick a variable and label everything clearly:age now:Hage two years from now:H+ 2age three years ago:H 3Now I need certain fractions of these ages:one-half of age two years from now: (1/2)(H+ 2) =H/2+ 1one-third of age three years ago: (1/3)(H 3) =H/3 1The sum of these two numbers is twenty, so I'll add them and set this equal to20:H/2+ 1 +H/3 1 = 20H/2+H/3= 203H+ 2H= 1205H= 120H= 24Heather is24years old.

Done by: waha Agalin Submitted to: teacher Desiree