Three Equations, Three Unknowns

$(x + y) + (y + z) + (z + x) = 10 + 20 + 24$

$2x + 2y + 2z = 54$

$2 (x + y + z) = 54$

$x + y + z = 27$

First add all the equations...

$(x+y)+(y+z)+(z+x)=10+20+24$

$2x+2y+2z=54$

Then you see that the two is common in the whole equation and divide the whole equation by two...

$2(x+y+z)=54$

$x+y+z=54/2$

$x+y+z=27$

And there is your answer...

if we add these three equation then

x+y+y+z+z+x=10+20+24

2x+2y+2z=54

2(x+y+z)=54

x+y+z= 54/2

x+y+z=27

If we add the three equations,

x + y = 10 (1).

y + z = 20 (2).

z + x = 24 (3).

Adding (1) (2) (3) together we obtain:

2x + 2y + 2z = 54

Thus, dividing by 2 we obtain:

x + y + z = 27

Hence the answer is 27

First

X + Y = 10

Y + Z = 20

Z + X = 24

Second

Z - Y = 14

Z + Y = 20

2 Z = 34

Third

Z = 17

Y = 3

X = 7

X+Y+Z = $\boxed{27}$

(x+y) +(y+z) +(z+x) =10+20+24 2(x+y+z)=54 x+y+z= 54/2 x+y+z=27

add all three formulas we get, 2x+2y+2z=54 taking 2 common we get x+y+z=54/2 Ans=27

x + y=10, y+z=20, z+x=24 add the three equations 2(x + y+ z)=54 x+y+z=27

x+y =10 -------(1) y+z = 20 -------(2) z+x = 24 -------(3)

From eqn (1) => x = 10 -y ------(4)

Replacing x in eqn (3) z+(10-y) = 24 z-y = 14 ---------(5)

Adding eqn (2) & (5) gives => 2z=34 which implies z = 17

Replace z in eqn (2) to find y value => y = 20-17 = 3

Replace y value in eqn (4) to find x value => x = 10-3 = 7

Hence the values of x,y,z are 7,3,17 So the answer to the given question (x+y+z) is 27

This is an easy one...

x = 10 - y

z + 10 - y = 24

z = 14 + y

y + 14 + y = 20

y = 3

z = 14 + 3 = 17

x = 10 - 3 = 7

x + y + z = 7 + 3 + 17 = 27

adding all these x+y+y+z+z+x=10+20+24 2(x+y+z)=54 x+y+z=54/2 x+y+z=27

using equ:1&3 i.e. (x+y = 10) - (z+x = 24) which gives, y-z = -14 adding this to equation (2) y-z = -14 + y+z =20 which gives 2y = 6 so y=3 , now adding y=3 to equ: 3 x=y=z = 24+3 = 27

simply x+y+z=27

Pick any equation out of 3, ie: I picked 2nd: y + z = 20. So I need to know the equation of y and z, which can be obtained from the question it self:

y = 10 - x

z = 24 - x

Hence:

   (10 -x) + (24 - x) = 20                                                                                                                                                                                             
   x = 7

So,
x + (y + z) = 7 + 20 = 27

7 + 3 = 10 3 + 17 = 20 17 + 7 = 24 so x = 7 y = 3 z = 17 x + y + z = 27

Add all three equations we get: 2x+2y+2z=10+20+24

After solving we get: x+y+z=27

take y+z = 20 ===> z = 20 - y,,, substitute the value z in z+x = 24 ====> 20 - y + x = 24===>x - y = 4,,, solve x + y = 10 and x - y = 4 ====> x = 7,,, substitute the value 7 in first equation ==> x+y = ,24 ===>y = 3,,, put y value in second equation ===>z = 17... x+y+z = 7+3+17=27.

-->(x+y)+(y+z)+(z+x)=10+20+24 -->2x+2y+2z=54 -->x+y+z=27 Ans: 27.

by adding x+y+y+z+z+x=54,2x+2y+2z=54,x+y+z=27.

Adding all three equation we get, (x + y) + (y + z) + (z + x) = 10 + 20 + 24 or, 2( x + y + z) = 54, Therefore (x + y + z) = 54/2 = 27

step1:- x=10-y, y+z=20, z+(10-y)=24;

step2:- x=10-y, y+z=20, z=14+y;

step3: x=10-y, 2y+14=20, z=14+y;

step4:- x=10-y, y=3, z=14+y;

step5:- x=7, y=3, z=17; ans:- x+y+z=27

x + y + y + z + z + x = 54. Then, 2 (x+y+z) = 54, and 54/2 = 27 Ans

x+y+y+z+z+x=10+20+24= 2x+2y+2z=54= 2(x+y+z)=54= x+y+z=54/2=27

Solution: X+Y=10…….(1) Y+Z=20………(2) Z+X=24………(3) Now, (1)+(2) Z+X+2Y=30 =>24+2Y=30 [From eqn. (3) ] So, Y=3 X=7 And Z=17 Now, X+Y+Z=27

When we add $x+y$ to $y+z$ we get $x+2y+z = 30$ . Now $x+z=30-2y$ . But we know that $x+z=24$ therefore we equate $30-2y$ with 24 and we get $y=3$ . With this information we can find the value of x and later the value of z. The answer is 27

adding three equations we have,

2x + 2y + 2z = 54

or 2(x+y+z) = 54

or x + y + z = 27

Adding the three equations we get 2(x+y+z)=24+10+20 = 54 Therefore x+y+z=54/2 = 27 Ans. 27

solve the equations for x,y, and z and find the result as27

x+y=10....(7,3) y+z=20....(3,17) z+x=24...(17,7) hence...x+y+z=24

Add all of equation

x+y+y+z+z+x =10 +20 +24

2x+2y+2z =54

taking 2 as Common

2(x+y+z)=54

Divide 2 on b/s

x+y+z=27;

by solving v get x=7 y=3 z=17 soo x+y+z=27

from x+y=10 y=10-x this in eq 2 we get 10-x+z=20 now from eq 3 z=24-x now 10-x+24-x=20 so x=7 substitute value of x in other eq we get y=3 & z=17 so........... x+y+z=27

X+Y+Y +Z +Z+X = 10+20+24 2(X+Y+Z) = 54 X+Y+Z=54/2 X+Y+Z =27

5+10+12=27

there in first equation x+y=10 so y=10-x.....if we put this value of y in second equation at the place of y then we get 10-x+z=20 .then z=10+x..then put this value of z in third equation then we get x=7...then put this value in first equation then we get y=3 ....after tht we get z=17..then x+y+z=27...

x+y+y+z+z+x=24+20+10 \ 2x+2y+2z+54 divide all with 2 \ x+w+z=27

Just add above three equations you will get 2x+2y+2z = 10+20+24 => 2(x+y+z) = 54 => x+y+z = 54/2 => x+y+z = 27

x+y =10 ........ y+z=20 ....... z+x=24 ..... adding 3 equations we get ...... x+y+y+z+z+x=10+20+24 ...... 2x+2y+2z=54 ,,,,, 2(x+y+z) =54 .... x+y+z=54/2 ... x+y+z=27

x= 3 y= 7

z= 17

answer: x + y + z = 27

x+y-y+z=10-20;
x-z=-10:
x-z+z+x=-10+24:
x=7:
y=3:
z=17:
7+3+17=27:

x+y=10 y+z=20 z+x= 24 5+5=10 10+10=20 12+12 = 24

x+y+z = 27 5+10+12=27

as x+y = 10 ,y=z =20 , z+x = 24 BY adding all the three equation we get 2x+2y+2z =54 by taking 2 as common we get 2(x+y+z) = 54 this implies, x+y+z= 27

Y+Z=20 *-1 =(-Y-Z=-20) +(Z+X=24)= (X-Y=4) +(X+Y=10)=2X=14 (X=7) (Y=3) ( Z=17) ( X+Y+Z)=7+3+17=24

adding all the three equations we get.. x+y+y+z+z+x=10+20+24

2x+2y+2z=54

we get 2(x+y+z)=54

x+y+z=27

x+y=10 y+z=20 x+z=24

                       x+y+y+z+x+z=10+20+24
                                2(x+y+z)=54

                                  x+y+z=27

add all these three equations we get 2x+2y+2z=54
2(x+y+z)=54 x+y+z=27

x+y=10 y+z=20 z+x=24

x=10-y z=24-10+y z-y=14 z+y=20 2z=34 z=17 x=7 y=3 x+y+z=7+3+17=27

x+y+y+z+z+x=10+20+24 2x+2y+2z=54 2(x+y+z)=54 x+y+z=54/2 =27

We just add x+y, y+z and z+x, Like x+y+y+z+z+x=10+20+24 We get 2x+2y+2z=54 Taking 2 common- 2(x+y+z)=54 Dividing 54/2 we get 27 so, x+y+z=27 Cheers!

If all the equations are added up, you can see than 2x + 2y + 2z = 10 + 20 +24 And thus divide both sides by two, and you get 27.

X+Y+Y+Z+Z+X=10+20+24 2X+2Y+2Z =54 2(X+Y+Z)=54 X+Y+Z=54/2 X+Y+Z=27

2(x+y+Z)=20+10+24 x+y+Z=54/2 =27

Here, x + y = 10 (...1)

y + z = 20 (...2)

z + x = 24 (...3)

Adding (...1), (...2) and (...3), we get,

(x + x) + (y + y) + (z + z) = 54

2(x + y +z) = 54

x + y + z = \frac {54}{2}

x + y + z = 27

• x+y=10 (x in terms of y) x= 10-y *y+z=20 (z in terms of x) z=20-y *z+x=24 substitute by y 20-y+10-y=24 20+10-24=2y divide by 2 y=3 Then x=7 and z=17 *x+y+z= 3+7+17=27

x+y+y+z+z+x=10+20+24 2x+2y+2z=54 2(x+y+z)=54 x+y+z=27

add all equations...... now u will see the pattern..

I solved this problem replacing the unknowns in the equations. So, if x + y = 10, y = 10 - x. So, if 10 - x + z = 20, z = 20 - 10 + x. Replacing in the last equation, 20 -10 + x + x = 24. Solving, 2x = 14, and x = 7. Replacing again, y = 10 - 7, so y = 3 and z = 10 + 7, so z = 17. Finishing, x + y + z = 7 + 3 + 17 = 27.

First you need to find out what x is by check through each number added to another number makes ten the see if it works for the other equations.

its very easy..... X+Y=10------------(1) Y+Z=20------------(2) Z+X=24------------(3) add three equations then we get 2(X+Y+Z)=54 implies X+Y+Z=27

x+y+0z=10 eq(1) 0x+y+z=20 eq(2) x+0y+z=24 eq(3) from eq 1 and 2 x-z=-10 eq(4) and from eq 3 and 4 x=7 x+z=24, 7+z=24. z=24-7=17 and eq(1) x+y=10 7+y=10, y=10-7=3

x+y+y+z+x=10+20+24 2x+2y+2z=54 2(x+y+z)=54 (Dividing both side by 2) x+y+z=27

just sum the three equations & then we get 2( x+y+z) = 54 and it implies that x+y+z = 27.

$(x + y ) + (y + z) + (z + x) = 10 + 20 + 24$

Then $2( x + y + z ) = 54$

$x + y + z = 27$

x+y=10

solve for y

y=10-x

(10-x)+z=20

solve for z

z=10+x

(10+x)+x=24

solve for x

x=7

y=10-7=3

z=10+7=17

x + y=10 z + x=24 =) x=10 - y --------(1) =) z=24-x-------------(2)

Put value of x nd z in y+z=20 we get 10-y+24-y=20 =) y= 3

put value of y in (1) nd (2) we get x= 7 nd z=17

Very easy!

Just sum the 3 lines and find 2x+2y+2z = 54. So, X + Y + Z = 54/2 = 27.

1. (x+y)+(y+z)+(z+x)=10+20+24
2. 2x+2y+2z=54
3. 2(x+y+z)=54
4. x+y+z=54/2
5. thus, x+y+z=27!

equation (1): x + y = 10 >>> y = 10 - x equation (1.1) equation (3): z + x = 24 >>> z = 24 - x equation (3.1) equation (2): y + z = 20 >>> substitute the value of y and z from equation (1.1 and 3.1) so, (10 - x) + (24 - x) = 20 >>> -2x + 34 = 20 >>>-2x = -14 >>> x = 7 using x = 7, find y and z equation (1) x + y = 10 >>> y = 10 - 7 >>> y = 3 equation (3) z + x = 24 >>> z = 24 - 7 >>> z = 14

x + y + z = 7 + 3 + 17 = 27

x+y+y+z+z+x=10+20+24 2x+2y+2z=54 2(x+y+z)=54 x+y+z=54/2 x+y+z=27

x=7, y=3 & z=17

get z's values from last 2 given equations ; compare both and then put in required equation , x's value which you got and y+z is given then we solve our problem by getting answer 27.

x=7,y=3, and z=17
7+3+17=27

First you figure out the different possibilities for x+y=10. Then, you figure out the different possibilities for y+z=20. After that, you see which pair has the same number for y. That means you know what x and z are because they are part of the first two equations and you already know what y is. You can check it over by doing the last equation and seeing if it is correct. You get x=7, y=3, and z=17. When you add them all together you get 27.

Adding we get x+y+y+z+z+x=10+20+24=54 2(x+y+z)=54 x+y+z = 54/2 = 27

x + y = 10 hence : y = 10 - x y + z = 20 hence : 10- x + z = 20 hence : z = 20 - 10 + x hence: z = 10 + x z + x = 24 hence : (10+x) + x = 24 hence : 10 + 2x = 24 hence : 2x = 14 x = 7 y = 3 z = 17 x + y +z = 7+3+17 = 27

I use the trial and error method. And I get X=7 Y=3 Z=17

First, take two of the three equations, x+y=10 and x+z=24. Using elimination, we end up with z-y=14. Now, take the untouched equation, z+y=20, and use elimination with the new equation. We end up with z=17. After plugging in z, we can find x and y, which are 7 and 3, respectively. Add 17, 7, and 3 to get an answer of 27.

Adding x+y,y+z and z+x, we get

x+y+y+z+z+x = 10+20+24

2(x+y+z) = 54

x+y+z = 54/2

x+y+z = 27

x+y=10 and y+z=20 solve these two equations linearly and find x-z and then solve those two equations then you get the final answer as 27

Add all the equation: 2(x+y+z) = 10+20+24 ; 2(x+y+z) = 54 ; x+y+z = 54/2 ; x+y+z = 27

2(x+y+z)=54 x+y+z = 27

FIRST OF ALL.. AS.. (x+y)+(y+z)+(z+x)=10+20+24=54 now, we will open the brackets and then solve this question as an equation. x+y+y+z+z+x=54 this implies that 2x+2y+2z=54 now as 2 is common: so we will take 2 outside the bracket.. 2(x+y+z)=54 now we will follow the rule of transposition.. this implies that(x+y+z)=54/2=27...

add 3 equations, we get 2(x+y+z)= 54.so x+y+z= 27

(x+y)+(y+z)+(z+x)=10+20+24 x+y+y+z+z+x=54 2x+2y+2z=54 take 2 as common 2(x+y+z)=54 2/2(x+y+z)=54/2 x+y+z=27

2(x+y+z)=54 so, x+y+z=27

x + y =10...(1) y + z =20...(2) z + x =24...(3) x + y + z =?? ...(4)

Add eq.(2) and eq.(3) x + y + 2z = 44 from eq.(1) we know that x+y=10 10 + 2z = 44 2z = 34 z = 17. Put x+y=10 and z=17 in eq.(4) x+y+z = 10+17 = 27.

x+y+y+z+z+x=10+20+24 x+y+z=27

x+y+y+z+z+x=10+20+24 2(x+y+z)=54 x+y+z=27

x+y+y+z+x+z=10+20+24 2x+2y+2z=54 2(x+y+z)=54 x+y+z=27

Q : x+y=10 A : 7+3=10

Q : Y+Z=20 A : 3+ 17=20

Q : Z+X=24 A : 17+7=24

To add the values : x+y+z :
7+3+17= 27

by adding these 3 equations

x+y+y+z+z+x=10+20+24

2x+2y+2z=54

2(x+y+z)=54

(x+y+z)=(54/2)

(x+y+z)=27

x +y = 10 (i) y + z = 20 (ii) z + x = 24 (iii)

subt i from ii we get z - x = 10 add this to iii we get z = 17 put z = 17 in ii and iii we get y = 3 and x= 7 then x + y +z = 7 + 3 + 17 = 27

Add all 3 equations, then you get 2(x+y+z)=54; then x+y+z=54/2=27.

2x+2y+2z=54. x+y+z=27

x+y=10
y+z=20
z+x=24
so

find the value of x
x+y=10 ------> x=10-y
convert it to the third equation because in the second equation there is no x
z+x=24 ------> z+10-y=24 find the value of z so that we can use it to the second equation so z=14+y
convert now the value of z to the second equation
y+z=20 ----> y+14+y=20 ----combine like terms----> 2y=6 where y=3
use the value of y to get x and z
until you find the answer where x=7 , y=3 , z=17 so x+y+z=27 ('_')

X =10-Y , Y=20-Z , Z=24-X ,

YOU NEED TO MAKE SURE THERE IS ONLY ONLY 1 UNKNOWN.I CHOOSE Z TO BE THE UNKNOWN BY CHOOSING AN EQUATION AMONG THE 3 EQUATIONS TO SUBSTITUTE OTHER DERIVED EQUATIONS TO MAKE SURE THERES ONLY ONE UNKNOWN, Z.

I CHOSE EQUATION X=10 - Y TO SUBTITUTE OTHER EQUATIONS

Z=24-X IS MODIFIED TO X=24-Z

NEXT SUSTITUTE THE EQUATIONS

24-Z=10-(20-Z) , Z=17 , Y=20-Z , y=3 , x=10-y , x=7 , x+y+z=7+3+17 , X+Y+Z=27

Adding all the 3 we get 2(x+y+z)=54 so x+y+z=27

x+y+y+z+z+x=10+20+24 2x+2y+2z=54 then, divide both sides by 2 x+y+z=27

(y+z) - (x +y)= (z-x); (z+x) +(z-x) gives d value of z. putting z in 2nd eq. solves y and samely replacing y's in 1st eq gives x. all of the variables can be added then.