My Calculator Story

In the story.

$x + \frac{y}{z} = 12$ ----> eq 1

$y + \frac{x}{z} = 9$ ----> eq 2

$\frac{x+y}{z} = 7$ ----> eq 3

Add eq 1 and eq 2

$x+y+ \frac{x+y}{z} = 21 \Rightarrow x+y +7 = 21$

So, $x+y = 14$

By eq 3, $\frac{x+y}{z} = 7 \Rightarrow \frac{14}{z} = 7$

So, $z = 2$ . Therefore $x+y+z = 14+2 = 16$ .

Note that this problem is edited

x + y = 7z

1) zx + y = 12z

2) yz + x = 9z

Add 1) and 2)

zx + zy + x + y = 21z --> zx + zy + 7z = 21z--> z(x+y+7) = 21z --> x + y = 14

So, 7z = 14 --> z = 2

Finally, x + y + z = 16

the solution will be have 2 answer if you using eq 1 and eq 2 substitute to eq 3 example eq 1 => x + (y/z)= 12 x = 12 - (y/z)

eq 2 => y + (x/z)= 9 y = 9 - (x/z)

then substitute to eq 3 = > ( x + y )/z = 7

[12 - (y/z) + 9 - (x/z)] / z = 7 12 - (y/z) + 9 - (x/z) = 7z 21 - (y/z) - (x/z) = 7z multiply all with z 21z - y - x = 7z^2 0 = 7z^2 - 21z + (x+y)

finding (x+y) by using eq 1 and eq 2 then eq 3 x + y + (y/z) + (x/z) = 12 + 9 x + y + [(x+y)/z] = 21 x + y + 7 = 21 x + y = 14

then

0 = 7z^2 - 21z + 7 0 = (7z - 14) (z - 1) z = 1 or 7z = 14 >>> z = 2 x + y + z = 14 + 1 = 15 or x + y + z = 14 + 2 = 16

x+ $\frac{y}{z}$ =12 (eq.1)

y+ $\frac{x}{z}$ =9 (eq.2)

$\frac{x+y}{z}$ =7

x+y=7z (eq.3)

ADD EQ.1 AND EQ.2

x+y+ $\frac{x+y}{z}$ =21

SUB IN EQ.3

7z+7=21

z+1=3

z=2

SUB IN Z=2 INTO EQ.3

x+y=14

Therefore:

x+y+z=14+2

x+y+z=16

x+y/z=12

y+x/z=9

So adding these two we have x+y=14

Now we know from question (x+y)/z=7

So z=2

Thus x+y+z = 14+2=16

$z$ cannot be 1 because that would yield $x+y=12$ and $y+x=9$ . All of the number are relatively low integers, so try z=2.

By replacing $z$ as 2 and doubling to remove fractions, last equation becomes x+y=14 and first equation is 2x+y=24

Then $x=10, y=4, z=2$ by simple substitution or subtraction. Sum x, y, and z to get 16.

If x+y/z = 12, and y+x/z = 9, y must be 3 multiples of z larger than x. Next, (x+y)/z = 7. Here we can tell that x and y together are the seventh multiple of 7. We can take x as 0, 1, 2, or 3,(position of multiple) with y corresponding as 7, 6, 5 or 4 respectively(again, the position of multiple). only 2 and 5 of the corresponding multiple additions have a difference of 3. Now we have to find out what z is to solve the problem, which I, as a lazy person, would prefer to list it all down. To my luck, it was a very small number 2. There is probably a mathematical way to calculate for 2 after using my method, but as I said, I am extremely lazy.

Let x+y/z=12.........(1) . y+x/z=9..........(2).(x+y)/z=7 or,x/z+y/z=7..........(3). Now by doing (Eq1+Eq2-Eq3), we get x+y=14. By Eq3, 14/z=7 and we get z=2.Finally, x+y+z=14+2=16.

3 equations, 3 unknowns. With established values for z and x + y, we can plug in and solve for x and y: x = 10, y = 4, and as prev. z = 2.

x+(y/z)=12

y+(x/z)=9

(x/z)+(y/z)=7

xz+y+yz+x+x+y=28z

xz+yz+2x+2y=28z

2x+2y=28z-xz-yz

2(x+y)=z(28-x-y)

2(x+y)/z=28-x-y

14=28-x-y

x+y=14

14/z=7

z=2

14+2=16

Because the solutions are such low numbers, you can assume z=2 first. Therefore, x+y has to equal 14 (because (x+y)÷z (or 2) =7). You also will be able to assume x and y are even numbers because both will need to be divisible by 2. Start with your option 6 and 6. Well you get separate answers for each being divided by z so that isn't it. Try x=8 and y=4. 4÷2+8=10. You need to add 2 to get the solution to be 12, so increase x by 2. Now x=10, y=4. Equation 1 checks out. Now do 10÷2+4. You get 9. Equation 2 checks out. now do (10+4)÷2. Answer is 14. So x=10, y=4, z=2.

Always guess and check

Getting rid of fractions in each equation leaves: xz + y = 12z yz + x = 9z x + y = 7z Add 1st two equations: xz + x + yz + y = 21 z Factor left side and substitute on right: x(z+1)+ y(z+1)=3 (7z)= 3(x + y) Factor again and simplify (x+y)(z+1) = 3(x+y) z+1 =3 ==> z =2 Substitute into first equation: 2x+ y =24 ===> x + x +y = 24 Substitute: x + 14 = 24 ===> x = 10 so 2(10) + y =24 ===> y =4 x + y + z = 16