Could Someone Help? 5/7 + 5/6?

Mathematics

2 answers:

lara31 [8.8K]3 years ago

7 0

Answer: 5/42

Step-by-step explanation:

lakkis [162]3 years ago

5 0

Answer:

So the answer is:

Step-by-step explanation:

Here's how to subtract 5/7 from 5/6:

Step 1

We can't subtract two fractions with different denominators. So you need to get a common denominator. To do this, you'll multiply the denominators times each other... but the numerators have to change, too. They get multiplied by the other term's denominator.

So we multiply 5 by 7, and get 35.

Then we multiply 5 by 6, and get 30.

Next we give both terms new denominators -- 6 × 7 = 42

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Only problem 6a. Can someone demonstrate how to do this so I can do the rest solo?

Lisa [10]

as far as I can tell, is simply asking to write two more expressions, that are equivalent to the provided one, namely, grab the provided one and expand it, if you simplify the expanded version, you'd end up with the provided, for example

$\bf \boxed{6a.1}~\hfill \stackrel{changing}{\cfrac{29\cdot 3}{30\cdot 3}}\implies \stackrel{one}{\cfrac{87}{90}}~\hfill \stackrel{changing}{\cfrac{29\cdot 7}{30\cdot 7}}\implies \stackrel{t wo}{\cfrac{203}{210}}\\\\\\\boxed{6a.3}~\hfill \stackrel{ch anging}{\cfrac{15\div 3}{30\div 3}}\implies \stackrel{one}{\cfrac{5}{10}}~\hfill \stackrel{changing}{\cfrac{15\div 5}{30\div 5}}\implies \stackrel{two}{\cfrac{3}{6}}$

so let's do 6a1, 6a3 and 6a5.

$\bf \boxed{6a.1}~\hfill \stackrel{changing}{\cfrac{29\cdot 3}{30\cdot 3}}\implies \stackrel{one}{\cfrac{87}{90}}~\hfill \stackrel{changing}{\cfrac{29\cdot 7}{30\cdot 7}}\implies \stackrel{t wo}{\cfrac{203}{210}} \\\\\\ \boxed{6a.3}~\hfill \stackrel{changing}{\cfrac{15\div 3}{30\div 3}}\implies \stackrel{one}{\cfrac{5}{10}}~\hfill \stackrel{changing}{\cfrac{15\div 5}{30\div 5}}\implies \stackrel{two}{\cfrac{3}{6}}$

$\bf \boxed{6a.5}~\hfill \stackrel{changing}{(9\cdot 10)\div (2\cdot 10)}\implies \stackrel{one}{90\div 20}~\hfill \stackrel{changing}{(9\cdot 70)\div(2\cdot 70)}\implies \stackrel{two}{630\div 140}$