How To Add Fractions With Letters
23
Calculation ALGEBRAIC FRACTIONS
Different denominators -- The LCM
2nd level
THere IS One RULE for calculation or subtracting fractions: The denominators must be the same -- just every bit in arithmetic.
| a c | + | b c | = | a + b c |
Add together the numerators, and place their sum
over the common denominator.
| Example 1. | 6x + three 5 | + | 4x − 1 5 | = | tenx + 2 5 |
The denominators are the same. Add together the numerators as like terms.
| Example ii. | vix + 3 5 | − | 4ten − 1 five |
To subtract, modify the signs of the subtrahend, and add.
| sixx + 3 5 | − | iv10 − 1 five | = | 610 + 3 − ivx + 1 5 | = | 210 + 4 5 |
Problem 1.
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| a) | ten 3 | + | y 3 | = | 10 + y three | b) | 5 ten | − | 2 x | = | three x |
| c) | x x − 1 | + | ten + one x − i | = | two10 + ane ten − ane | d) | 3x − 4 x − 4 | + | x − 5 x − 4 | = | ivten − 9 x − four |
| e) | vix + i ten − 3 | − | 4x + 5 x − 3 | = | 6x + ane − 4ten − 5 ten − iii | = | iix − iv x − 3 |
| f) | iix − three x − 2 | − | x − four x − 2 | = | 2x − 3 − 10 + 4 x − 2 | = | x + 1 x − 2 |
Different denominators -- The LCM
To add fractions with dissimilar denominators, we must learn how to construct the Lowest Common Multiple of a series of terms.
The Lowest Common Multiple (LCM) of a series of terms
is the smallest product that contains every cistron of each term.
For case, consider this series of three terms:
pqprps
We volition at present construct their LCM -- factor past factor.
To begin, information technology will have the factors of the first term:
LCM = pq
Moving on to the second term, the LCM must take the factors pr. Just it already has the factor p -- therefore, we demand add merely the factor r:
LCM = pqr
Finally, moving on to the final term, the LCM must contain the factors ps. But over again information technology has the factor p, so we need add only the factor s:
LCM = pqrs.
That product is the Lowest Mutual Multiple ofpq, pr,ps. It is the smallest product that contains each of them as factors.
Example three. Construct the LCM of these three terms:x,x 2,x iii.
Solution. The LCM must take the cistron ten.
LCM = 10
But information technology likewise must take the factors of ten 2 -- which are x · x. Therefore, we must add together one more than factor of ten :
LCM = x 2
Finally, the LCM must have the factors of 10 three, which are 10 · x · 10. Therefore,
LCM = x 3.
x 3 is the smallest product that contains x,x ii, and10 3 equally factors.
Nosotros see that when the terms are powers of a variable -- ten,x 2,x three -- so their LCM is the highest power.
Problem 2. Construct the LCM of each series of terms.
| a) | ab,bc,cd. abcd | b) | pqr,qrs,rst. pqrst | |
| c) | a,a 2,a 3,a four. a iv | d) | a 2 b,a b 2. a 2 b 2 | |
east)ab,cd. abcd
Nosotros will now encounter what this has to do with calculation fractions.
| Example 4. Add: | 3 ab | + | 4 bc | + | v cd |
Solution. To add fractions, the denominators must exist the same. Therefore, as a common denominator cull the LCM of the original denominators. Cull abcd. And then, convert each fraction to an equivalent fraction with denominator abcd.
It is necessary to write the mutual denominator only one time:
| 3 a b | + | 4 b c | + | 5 c d | = | 3 c d + iv a d + 5 a b a b c d |
To modify 
into an equivalent fraction with denominator a b c d , but multiply a b by the factors information technology is missing, namely c d . Therefore, nosotros must also multiply 3 by c d . That accounts for the start term in the numerator.
To change 
into an equivalent fraction with denominator a b c d , multiply b c by the factors it is missing, namely a d . Therefore, we must also multiply 4 past a d . That accounts for the 2d term in the numerator.
To change 
into an equivalent fraction with denominator a b c d , multiply c d by the factors it is missing, namely a b . Therefore, we must also multiply v past a b . That accounts for the terminal term in the numerator.
That is how to add fractions with different denominators.
Each gene of the original denominators must be a gene
of the mutual denominator.
Problem 3. Add together.
| a) | 5 ab | + | 6 ac | = | vc + sixb abc |
| b) | two pq | + | iii qr | + | four rs | = | 2rs + 3ps + 4pq pqrs |
| c) | 7 ab | + | eight bc | + | nine abc | = | 7c + 8a + 9 abc |
| d) | i a | + | two a 2 | + | 3 a 3 | = | a 2 + 2a + iii a three |
| e) | three a 2 b | + | 4 a b 2 | = | 3b + iva a 2 b 2 |
| f) | 5 ab | + | vi cd | = | 5cd + half-dozenab abcd |
| g) | _2_ ten(x + two) | + | __3__ (10 + ii)(x − iii) | = | two(x − iii) + iiix x(10 + 2)(ten − three) |
| = | _ 2x − half-dozen + threex_ ten(10 + 2)(x − iii) | ||||
| = | _5x − 6_ 10(x + 2)(x − 3) | ||||
At the 2d Level we will encounter a similar trouble, but the denominators will not exist factored.
| Problem 4. Add: ane − | i a | + | c + 1 ab | . Only write the answer equally |
1 − a fraction.
| 1 − | 1 a | + | c + one ab | = | 1 − ( | 1 a | − | c + 1 ab | ) |
Case 5. Denominators with no common factors.
When the denominators have no common factors, their LCM is simply their product, mn.
| a m | + | b northward | = | an + bm m n |
The numerator and then appears equally the effect of "cross-multiplying" :
an + bm
However, that technique volition piece of work simply when calculation 2 fractions, and the denominators have no common factors.
| Example vi. | ii x − one | − | 1 x |
Solution. These denominators take no mutual factors -- 10 is not a factor of x − ane. Information technology is a term. Therefore, the LCM of denominators is their product.
| 2 10 − 1 | − | ane x | = | 2ten − (x − 1) (x − 1)x | = | 2x − x + 1 (ten − one)x | = | _x + 1_ (x − 1)ten |
Annotation: The entirex − one is being subtracted. Therefore, we write it in parentheses -- and its signs alter.
Problem 5.
| a) | x a | + | y b | = | xb + ya ab | b) | x five | + | 3ten two | = | twoten + 1510 10 | = | 17x 10 |
| c) | 6 x − i | + | iii x + i | = | vi(x + 1) + 3(10 − i) (x + one)(x − 1) |
| = | 6ten + 6 + threex − 3 (10 + 1)(x − 1) | ||||
| = | _9x + 3_ (x + one)(x − 1) | ||||
| d) | 6 10 − ane | − | 3 x + 1 | = | 6(10 + 1) − three(x − 1) (x + 1)(ten − 1) |
| = | sixx + 6 − 310 + 3 (x + one)(ten − i) | ||||
| = | _3x + 9_ (ten + i)(10 − 1) | ||||
| e) | 3 ten − 3 | − | two x | = | 3x − 2(x − 3) (x − 3)x |
| = | 3x − 2x + half-dozen (x − three)x | ||||
| = | x + half-dozen (x − iii)10 | ||||
| f) | iii x − 3 | − | 1 x | = | 3x − (x − iii) (x − 3)x |
| = | 3x − 10 + iii (x − 3)x | ||||
| = | 2x + 3 (x − 3)x | ||||
| g) | 1 x | + | 2 y | + | 3 z | = | yz + 2xz + 3xy xyz |
| Instance 7. Add:a + | b c | . |
Solution. Nosotros have to express a with denominator c.
Therefore,
| a + | b c | = | air conditioning + b c | . |
Problem 6.
| a) | p q | +r | = | p + qr q | b) | 1 ten | − 1 | = | i − x ten |
| c)x − | 1 x | = | 10 2 − 1 x | d) 1 − | i x 2 | = | x two − 1 10 two |
| e) 1 − | 1 x + one | = | x + one − 1 x + 1 | = | x x + ane |
| f) three + | 2 x + one | = | 3x + three + 2 x + i | = | three10 + five x + 1 |
| Problem vii. Write the reciprocal of | 1 2 | + | i three | . |
| [Hint: But a single fraction | a b | has a reciprocal; it is | b a | .] |
| one 2 | + | ane 3 | = | 3 + ii half dozen | = | 5 6 |
| Therefore, the reciprocal is | half-dozen 5 | . |
2nd Level
Next Lesson: Equations with fractions
Table of Contents | Dwelling house
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How To Add Fractions With Letters,
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