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six times a number is subtracted from eight times its reciprocal. The result is 47. Find the number

six times a number is subtracted from eight times its reciprocal. The result is 47. Find the number

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eight times the reciprocal of a number equals 2 times the ...


Three times the reciprocal of a number equals 9 times the reciprocal of 6. Find the number. ... number equals 47 and 10 times ... reciprocal of a number , the result ...
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Three times the reciprocal of a number equals 9 times the ...


The sum of a number and 20 times its reciprocal is 9. Find the number ... 3 times the second number equals 47 and 10 ... number is subtracted from six times the ...
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Three times the reciprocal of a number equals 9 times the ...


Three times the reciprocal of a number equals 9 times the reciprocal of 6 ... times a number is subtracted from ... times its reciprocal the result is 14 find ...
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TRANSLATE WORD SENTENCES INTO ALGEBRAIC EXPRESSIONS


... Eight subtracted from two times a number. ... The sum of a number and its reciprocal is equal to four. ... TRANSLATE WORD SENTENCES INTO ALGEBRAIC ...
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less than or subtracted from - Mt. San Antonio College


greater than ratio of the result is ... “less than ” or “subtracted from ”. ... Twice a number less than four is equal to six times a number plus five. 2 ...
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PRACTICE PROBLEMS INVOLVING RATIONAL EXPRESSIONS


PRACTICE PROBLEMS INVOLVING RATIONAL EXPRESSIONS. ... The sum of a number and its reciprocal is 17/4. Find the ... times a number is subtracted from twice its ...
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Translating Words into Mathematical Symbols


Translating Words into Mathematical Symbols ... The sum of 4 times a number n and 7 4n + 7 ... sum The result of adding numbers
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Translating Verbal Expressions – Terms - Del Mar College


Translating Verbal Expressions – Terms . ADDITION: ... The difference of four times a number and six 4x – 6 ... 2/6/2017 8:47:06 AM ...
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Suggested Questions And Answer :


six times a number is subtracted from eight times its reciprocal. The result is 47. Find the number

me thank yu tri tu sae (8/x) -6x=47 8 -6x^2-47x=0 or 6x^2 +47x-8=0 quadratik equashun giv roots=0.16666666, & -8
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if eight times a number is subtracted from eleventimes the number,the result is -9.find the number.

This is P.ANDRONICUS so sumprobulums group 4 results so I forgot hall ticket nember so best answers tell me know so I will reqist sir&madam.
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how do i add fractions

To add or subtract fractions, obtain a least common denominator. Subtract the numerators in the correct order and retain the same least common denominator for your answer. Simplify. To multiply fractions, multiple the numerators. The product will be the numerator of your answer. Repeat with denominators. Simplify. To divide fractions, take the reciprocal of what you are dividing by. Multiply the reciprocal with the initial number (see above for multiplication process). Simplify. Evaluate means to solve. You can solve fraction problems using the above processes. You can only simplify if both the numerator and denominator are divisible by the same number. If the denominator is odd, you can only simplify it if the numerator also is divisible by a same number. Ex. 88/33. Although the denominator is odd, both the numerator and denominator are divisible by 11 resulting in 8/3 as the simplified answer. To pace yourself during a test do the following. Find out how long you have for the test. Divide this by the total number of problems on the test. Example. 1 hour for 20 problems on your test. This means you have 3 minutes per problem. If you spend more than 3 minutes on a problem, skip it. Continue until you attempt all the problems. Go back with the remainder of the time to retry these problems you skipped. Most likely they are the most difficult, hence why you spent alot of time on them. This method of pacing allows you to skip the hard problems at first, attempt all problems, and finish the easier problems for sure.
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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hpw do you estimate an answer to a problem with a remainder?

The purpose of the estimate is not to get an accurate answer, but just an indication of what the answer should be approximately, so that when the actual answer is determined, you can compare it with the estimate. They should be fairly close. If they are, this gives you confidence in your ability to work out the answer. If they're not, you may have made a mistake in your calculations. With practice, you become used to estimating quickly and that gives you a rough expectation of what the answer should be. So when you're doing a division it's not the remainder you need to be concerned about, but rather a rough idea of what the answer is going to be when calculated thoroughly. For example, if you were asked to divide 490 by 21, you could quickly estimate this by using 500 in place of 490 and 20 in place of 21, because these substitute numbers are close enough, or compatible with the actual numbers. 500 divided by 20 is 25. That can be done quickly in your head. 490/21 is going to be roughly 25. The actual answer is 23 with a remainder of 7. The estimate doesn't try to predict the remainder, but 25 is close enough to 23 to tell you that the real answer is probably correct. If you calculated 490/21 to be something way different from 25, you would suspect something had gone wrong in your calculations. So that's it: the estimate is just a quick fire judgment, before you take the time to do the proper calculation. An estimate is, or should be a quick mental calculation and not a laborious or lengthy exercise. For some numbers there is a way of predicting the remainder without doing any division at all. The obvious one is dividing by 2, because all odd numbers will give remainder 1. Dividing by 4 requires dividing only the last two digits by 4 and noting the remainder. Dividing by 8 requires dividing the last three digits and noting the remainder. Dividing by 5 is just a matter of noting whether the number ends in zero or 5. If it does there's no remainder; if it doesn't the remainder is found by subtracting 5 or zero from the last digit, depending on the size of the digit. If the digit is between 1 and 4 then so is the remainder; if it's between 6 and 9 then the remainder is found by subtracting 5. When dividing by 9, we don't do any division at all: we simply add up the digits making the number and if the result is 10 or more we add the digits of the result and we keep doing this until we end up with a single digit. If this digit is 9 the number is divisible exactly by 9; otherwise the digit is the actual remainder (this method is not surprisingly called the 9's remainder, and it can be used to check addition, subtraction and multiplication with a 90% accuracy level). Division by 11 is slightly more tricky, but consists of, starting with the last digit, add the alternate digits of the number then subtract the sum of the remaining digits. If the result is positive then that is the remainder; if negative then add 11 to find the remainder; for example, to find the remainder of 98765 divided by 11, we add 5+7+9=21 and subtract 6+8=14: 21-14=7, so the remainder is 7. Another example: remainder of dividing 23451 by 11; 1+4+2-(5+3)=-1+11=10, so the remainder is 10. If the result of the alternate subtraction exceeds 11 repeat the process: 190827 gives 7+8+9-(2+0+1)=21 so the next step is 1-2=-1, add 11=10, so 10 is the remainder.
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Word problem, What is the answer, please show work!

if 4 is multiplied times the sum of a number and 6, the result is 28 less the number. what is the number?   As an equation i think this would be: 4(x + 6) = x - 28 4x + 24 = x - 28   Now you can subtract 24 frm both sides of the equation: 4x + 24 - 24 = x - 28 - 24 4x = x - 52   Now you can subtract x from both sides of the equation: 4x - x = x - 52 - x 3x = - 52   Now you can divide both sides of the equation by 3 to find the value of x on it's own: 3(x)/3 = - 52/3 x = - 52/3   So the number you wanted to find is -52/3
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Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42}

Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42} Please just solve the set provided above!!!! This will be a bit more involved than the systems with two unknowns, but the process is the same. The plan of attack is to use equations one and two to eliminate z. That will leave an equation with x and y. Then, use equations one and three to eliminate z again, leaving another equation with x and y. Those two equations will be used to eliminate x, leaving us with the value of y. I'll number equations I intend to use later so you can refer back to them. That's enough discussion for now. 1)  3x-2y+2z=30 2)  -x+3y-4z=-33 3)  2x-4y+3z=42 Equation one; multiply by 2 so the z term has 4 as the coefficient. 3x - 2y + 2z = 30 2 * (3x - 2y + 2z) = 30 * 2 4)  6x - 4y + 4z = 60 Add equation two to equation four:   6x - 4y + 4z =  60 +(-x + 3y - 4z = -33) ----------------------   5x - y       = 27 5)  5x - y = 27 Multiply equation one by 3. Watch the coefficient of z. 3 * (3x - 2y + 2z) = 30 * 3 6)  9x - 6y + 6z = 90 Multiply equation three by 2. Again, watch the coefficient of z. 2 * (2x - 4y + 3z) = 42 * 2 7)  4x - 8y + 6z = 84 Subtract equation seven from equation six.   9x - 6y + 6z = 90 -(4x - 8y + 6z = 84) ----------------------   5x + 2y      =  6 8)  5x + 2y = 6 Subtract equation eight from equation five. Both equations have 5 as the coefficient of x. We eliminate x this way.   5x -  y = 27 -(5x + 2y = 6) ---------------       -3y = 21 -3y = 21 y = -7  <<<<<<<<<<<<<<<<<<< At this point, I am confident that I followed the correct procedures to arrive at the value for y. Use that value to determine the value of x. ~~~~~~~~~~~~~~~ Plug y into equation five to find x. 5x - y = 27 5x - (-7) = 27 5x + 7 = 27 5x = 27 - 7 5x = 20 x = 4  <<<<<<<<<<<<<<<<<<< Plug y into equation eight, too. 5x + 2y = 6 5x + 2(-7) = 6 5x - 14 = 6 5x = 6 + 14 5x = 20 x = 4    same value for x, confidence high Proceed, solving for the value of z. ~~~~~~~~~~~~~~~ Plug both x and y into equation one. We will solve for z. Equation one: 3x - 2y + 2z = 30 3(4) - 2(-7) + 2z = 30 12 + 14 + 2z = 30 26 + 2z = 30 2z = 30 - 26 2x = 4 z = 2  <<<<<<<<<<<<<<<<<<< Continue using the original equations to check the values. Equation two: -x + 3y - 4z = -33 -(4) + 3(-7) - 4z = -33 -4 - 21 - 4z = -33 -25 - 4z = -33 -4z = -33 + 25 -4z = -8 z = 2   same value for z, looking good Equation three: 2x - 4y + 3z = 42 2(4) - 4(-7) + 3z = 42 8 + 28 + 3z = 42 36 + 3z = 42 3z = 42 - 36 3z = 6 z = 2  satisfied with the results We have performed several checks along the way, thus proving all three of the values. x = 4, y = -7 and z = 2
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how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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Five times the difference of twice a number and eight is negative ten. Find the number.

call the number you are looking for x times=multiply difference=subtract twice a number=times 2 is=Equals   5(2x-8)=-10   Solve this for x to get your answer 10x-40=-10 10x=30 x=3
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what is 5 1/8-3 2/3

  5 1/8-3 2/3  You first need to find the L.C.D. (Lowest Common Denominator), since you can't subtract with different denominators. The easiest way to do this is to multiply the denominators. 8 X 3 = 24 Let's start with 1/8. The denominator - 8 goes into 24 THREE times. 3 times the numberator - 1 equals 3. So, the first number becomes 5 3/24. Now, let's work with 2/3. 3 goes into 24 EIGHT times. 8 times 2 equals 16. So the second number becomes 3 16/24. Now the problem is 5 3/24 - 3 16/24. Normally, you just subtract the numerators, & solve the problem. BUT you can't subtract 16 from 3 in te numerators, so you need to add to 3/24. You take 1 from 5, which makes it 4. Then you change the 1 to the equivalent we can work with here, which is 24/24. Now we add 24/24 to 3/24 to get 27/24. Now the problem becomes: 4 27/24 - 3 16/24 NOW we can subtract the numerators: 27 - 16 = 11. then we subtract the whole numbers: 4 - 3 = 1 This gives us the answer of: 1 11/24
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