Question

Can someone help with these 2 problems please!

Answer 1

first one
\[\frac{a-1}{7}+\frac{a-2}{4}=\frac{6}{7}\]

Answer 2

get rid of the denominators
ie multiply by each term by 28
\[4\times({a-1})+7\times({a-2})=4\times{6}\]

Answer 3

expand brackets\[4a-4+7a-14=24\]and simplify\[3a-18=24\]add eighteen to both sides\[3a=6\]divide by three\[a=2\]

Answer 4

i made a mistake

Answer 5

\[4a+7a=11a≠3a\]

Answer 6

I think it works out to 11a=42, so a=42/11

Answer 7

i actually made two mistakes
AnimalAin is correct
\[24+18=42≠6\]\[\cdots \]a\[11a-18=24\]\[11a=42\]\[a=42/11\]

Answer 8

second one (ill try to be more careful)
\[\frac{3r-3}{r^2+6r-7}+\frac{3}{r+7}=\frac{3}{r-1}\]
first step is to factorize the denominator of the first term

Answer 9

\[{r^2+6r-7}=(r+...)(r-...)\]

Answer 10

so we need two numbers that have a sum of 6 and a product of -7

Answer 11

So it would be 7 and -1

Answer 12

correct

Answer 13

\[\frac{3r-3}{(r+7)(r-1)}+\frac{3}{r+7}=\frac{3}{r-1}\]
now we have to get rig of the denominators

to do this
multiply all terms by \[(r+7)(r-1)\]

Answer 14

Would the bottom ones cancel out ?