Algebra_2

$\begin{cases} \dfrac {a+b}c = \dfrac 65 & \implies 5a+5b = 6c & \implies -5a+6c = 5b & ...(1) \\ \dfrac {b+c}a = \dfrac 92 & \implies 2b+2c = 9a & \implies 9a-2c = 2b & ...(2) \end{cases}$

$(1)+(2): \quad 4a+4c = 7b \implies \dfrac {a+c}b = \boxed{\dfrac 74}$

$\frac{a+b+c}{c} = \frac{a+b}c + \frac c c = \frac 6 5 + 1 = \frac{11}5.$ $\frac{a+b+c}{a} = \frac{b+c}a + \frac a a = \frac 9 2 + 1 = \frac{11}2.$ $\frac{b}{a+b+c} = \frac{a+b+c}{a+b+c} - \frac{a}{a+b+c} - \frac{c}{a+b+c} = 1 - \frac 2{11} - \frac 5{11} = \frac 4{11}.$ $\frac{a+c}{b} = \frac{a+b+c}{b} - \frac b b = \frac{11}4 - 1 = \frac 7 4.$

The denominator of the first fraction is $5$ , so we plug in $c=5$ . Doing this gives $b=4$ and $a=2$ , which satisfies both equations. $\frac{a+c}{b}=\frac{2+5}{4}=\boxed{\frac{7}{4}}$

First we note that $a$ , $b$ and $c$ appear in the denominator and therefore cannot be zero. Let $\frac{a+c}{b}=x$ .

The three equations can be written in matrix form as $\underbrace{\begin{bmatrix} 1 & 1 & -6/5 \\ -9/2 & 1 & 1 \\ 1 & -x & 1 \end{bmatrix}}_{A(x)} \begin{bmatrix} a \\ b \\ c\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$

For the matrix equation to have a solution with nonzero $a$ , $b$ and $c$ , the matrix $A(x)$ must be singular. Then, solving the equation $\det A(x) = 0$ with respect to $x$ gives the result $x=\tfrac{7}{4}$ .

Since $\frac{a + b}{c} = \frac{6}{5}$ , $c = \frac{5a + 5b}{6}$ .

Since $\frac{b + c}{a} = \frac{9}{2}$ , $c = \frac{9a - 2b}{2}$ .

Since $c = \frac{5a + 5b}{6} = \frac{9a - 2b}{2}$ , $b = 2a$ .

Since $b = 2a$ and $c = \frac{5a + 5b}{6}$ , $c = \frac{5a + 5 \cdot 2a}{6} = \frac{5}{2}a$ .

Since $b = 2a$ and $c = \frac{5}{2}a$ , $\frac{a + c}{b} = \frac{a + \frac{5}{2}a}{2a} = \boxed{\frac{7}{4}}$ .

(5 being the denominator)Given that we have the number 5, we can suggest that b=4 and a=2.

1. c=5 + a(2)= 7
2. b(4)

7/4

Treat the equations as logical predicates: “If a+b=6 then c=5”, and “if a=2 then b+c=9”. Suppose c is 5 and a is 2, does this determine b consistently? Yes. Both equations require it to be 4.

My thoughts: a is even, since whatever a equals it can be reduced to 2. Therefore, since a+b=6 and a is even, so must b. There is only one answer with a even denominator so it must be correct.

First we eliminate a variable and rephrase the required fraction in terms of a and c alone.

(1) $\displaystyle b = \frac{6}{5}c - a = \frac{9}{2} - c$

(2) $\displaystyle \frac{a+c}{b} = \frac{a+c}{\frac{6}{5}c - a }$

Separating the variables for expression (1) gives the following: $\displaystyle \frac{11}{2}a = \frac{11}{5}c$

Substituting into (2) then gives: $\displaystyle \frac{\frac{7}{5}c}{\frac{4}{5}c}$

This gives the answer, $\displaystyle \boxed{\frac{7}{4}}$

$\frac{a+b}{c}$ = $\frac{6}{5}$ so a+b = 6 and c = 5. $\frac{b+c}{a}$ = $\frac{9}{2}$ so b+c = 9 and a = 2.

a = 2

a+b = 6

b = 4

c = 5

$\frac{a+b}{c}$ = $\frac{7}{4}$

$\frac{a+b}{c}$ = $\frac{6}{5}$ ...................(i) $\frac{b+c}{a}$ = $\frac{9}{2}$ ...................(ii) From (i) &(ii) We get, $\frac{a+b+b+c}{c+a}$ = $\frac{6+9}{5+2}$ OR, $\frac{c+a+2b}{c+a}$ = $\frac{15}{7}$ OR, 1+ $\frac{2b}{c+a}$ = $\frac{15}{7}$ OR, 2X $\frac{b}{c+a}$ = $\frac{8}{7}$ OR, $\frac{b}{c+a}$ = $\frac{4}{7}$ SO, $\frac{a+c}{b}$ = $\frac{7}{4}$

$(a+b):c:(a+b+c)=6:5:11,(b+c):a:(a+b+c)=9:2:11$ ,so $a:b:c=2:4:5,\frac{a+c}{b}=\frac{7}{4}$

(a+b = 6k) (c = 5k)

(b+c = 9T) (a = 2T)

(b = 9T - c) (b = 6k - a) From above equations we conclude: (T = K)

Then : (a+c = 5k + 2k = 7k) (b = 9k - 5k = 4k)

And by dividing them we get the answer 😃

( 7k/4k = 7/4 )

there is a limited number of guess and check for numbers adding up to 6, like 1 and 5, 2 and 4 etc, so its easy to guess and check

We can find the values for C and A because they are the denominators, and that's what we are dividing by. Therefore,

a+b/c = 6/5 → c=5

b+c/a = 9/2 → a=2

Now we do some algebra because we know a+b = c we can substitute the variables we know.

2 + b/5 = 6/5

2+ b = 6 b = 4

So that answer is 7/4 when we add at the last step.

There is a very quick solution without doing any algebra: the 4 in $\frac{7}{4}$ caught my attention and from the denominators of the fractions you can see immediately that a is even and c is odd. Furthermore you see that (a+b) is even and (b+c) is odd. (a+b) can only be even if b is even and also (b+c) is odd if and only if b is even. And in the solution options the only fraction with a even denominator is $\frac{7}{4}$ .

I took a slightly different approach:

$\frac{a+b}{c}$ = $\frac{6}{5}$

--> $a+b=\frac{6c}{5}$

--> $a+b+c=\frac{11c}{5}$

--> $\frac{5(a+b+c)}{11}=c$

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$\frac{b+c}{a}$ = $\frac{9}{2}$

--> $b+c=\frac{9a}{2}$

--> $a+b+c=\frac{11a}{2}$

--> $\frac{2(a+b+c)}{11}=c$

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Therefore, $a$ and $c$ together make up $\frac{7}{11}$ of the total sum of $a+b+c$

Then $b$ must make up the other $\frac{4}{11}$ of the sum of $a+b+c$

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So: $\frac{4(sum)}{11} = b$

--> $a+b+c=\frac{11b}{4}$

--> $a+c=\frac{7b}{4}$

--> $\frac{a+c}{b} = \frac{7}{4}$

Taking both initial statements as ratios, both add to 11 parts. a + b : c is 6 : 5 and b + c : a is 9 : 2. Taking c as 5 parts and a as 2 parts leaves b as 4 parts. This satisfies both conditions so (a + c)/b = (5 + 2)/4 = 7/4

The way I solved it includes almost no algebra here is how it works: $\frac{a+b}{c}$ = $\frac{6}{5}$ and $\frac{b+c}{a}$ = $\frac{9}{2}$ a + b makes an even number c is an odd number b + c makes an odd number their for the answers denominator must be even and the only answer with an even denominator is $\frac{7}{4}$

Divide both numerator and denominator by c in all the equations let $\frac{a}{c}$ =x and $\frac{b}{c}$ =y
we get x+y= $\frac{6}{5}$ and $\frac{y+1}{x}$ = $\frac{9}{2}$
Solve for x and y
x= $\frac{2}{5}$ and y = $\frac{4}{5}$
put x and y in 3rd equation and we get $\frac{a+c}{b}$ = $\frac{x+1}{y}$ = $\frac{7}{4}$

If u just assume denominator as correct values, value of b perfectly fits. Hence this is one of the solutions. Now finding any thing is simple

b+c=9 , c=5 , a=2 Therefore b=9-c=9-5=4 Substitute a=2 , b=4 , c=5 into (a+c)/b to give 7/4. You have to look at the problem as fractions

You guys might laugh at me, but it took me less than 30 seconds to solve the squestion. Here is how I did it. I looked at the information given, and thought to myself what if c = 5, then a+b will be 6. (Just look at the first piece of information given to us); Then I looked at the second bit of information, and thought to myself, if a = 2; then b+c will be = 9. And because, I have already assumed c = 5 ; then b will be 4.

So a = 2; b = 4; c = 5; And these values satisfy the information given to us.

The question is to find the value of a+c/b. So put in the values of a, b, and c. You will get 2+5/4 = 7/4

Bingo, I have a correct answer.

My 12 yo child chose the answer directly because 6+5=11 9+2=11 and 7+4=11 But I don't have any explanation!!! Who have? !

From the given 2 eq., eliminate b and generate a relation between a and c i.e. 5a=2c. Now put it in eq. 2(or in 1) to get relation between a and b( or b and c) i.e. 2a=b( or any else relations) Anyway, use these 2 relations in 3rd eq. to get the solution. Easy peasy!!

The simplest solution happens to work in this case, namely simply plugging in the values of the various numerators and denominators given, and solving for the 3 variables. Doing so gives a consistent answer in this case, which must then indeed be the ratio we're looking for. Now, sure, this isn't guaranteed to work (I think), as in that it probably won't always give a consistent system of values; but if it does provide a consistent answer, this simple method is entirely valid and thus there isn't any reason not to use it in that case.

So: assuming that $a + b = 6$ , $c = 5$ , $b + c = 9$ and $a = 2$ , then it follows from the 2nd and 3rd entry on that list that $b = 4$ . This is indeed consistent, as indeed $a + b = 2 + 4 = 6$ . So, these values are a valid solution. That means that the ratio being asked for, $\frac{a + c}{b} = \frac{2 + 5}{4} = \color{#20A900}\large{\boxed{\frac{7}{4}}}$

From the conditions, we derive a linear homogeneous system of equations where $R=\frac{a+c}{b}$ is the ratio we are looking for: $\begin{cases}\phantom{-}5a+5b-6c=0 \\ -9a+2b+2c=0 \\ \phantom{-5} a - Rb + c = 0\end{cases}$ We want a non trivial solution for the system so the determinant of the coefficient matrix must be zero: $\begin{vmatrix} 5 & 5 & -6 \\ -9 & 2 & 2 \\ 1 & -R & 1 \end{vmatrix}=77-44R=0\ \implies R=\boxed{\dfrac{7}{4}}$

$a + b = \dfrac{6}{5}\times c \implies c = \dfrac{5}{6}(a + b)$

Plugging into our second equation we get:

$\dfrac{b + \frac{5}{6}(a + b)}{a} = \dfrac{9}{2} \implies 27a = 6b + 5a + 5b \implies 11b = 22a \implies b = 2a$ .

Now plugging this into what we want to find:

$\dfrac{a + c}{b} = \dfrac{a + c}{2a} = \dfrac{1}{2} + \dfrac{1}{2}\times\dfrac{c}{a} = \dfrac{1}{2} + \dfrac{1}{2}\times\left(\dfrac{9}{2} - \dfrac{b}{a}\right) = \dfrac{1}{2} + \dfrac{1}{2}\times\left(\dfrac{9}{2} - 2\right) = \boxed{\dfrac{7}{4}}$ .

let $c=5k\\a=2k\\a+b=6k$ .

then $b=4k\\a+c=7k$ .

so the result was $\frac 7 4$