No es fácil no?

Question

X+x-y=5

Z+y+y=50

X+z+? =40

Z+z-y=25

Y+z+x=?

Answer 1

Juan, Es facil.

z + y + y = 50

z + z - y = 25

multiplica la primera línea por 2.

2z + 4y = 100

z + z - y = 25

restar la primera línea de la segunda.

5y = 75

divide ambos lados por 5

y = 15

Sustituye el valor 15 en las otras ecuaciónes en lugar de la variable y.

X + x -15 = 5

15 + x + z = ?

x + z + ? = 40

Sustituye el valor (15 + x + z) en la tercera ecuación en lugar de “?”.

x + z + 15 + x + z = 40

simplifica esta linea

2x + 2z + 15 = 40

recuerda la primera ecuación

X + x -15 = 5

simplifica esta linea

2x - 15 = 5

2x + 2z + 15 = 40

resuelve para x, sustituye en la segunda línea y resuelve para z.

Answer 2

I'm assuming that X=x, Y=y, Z=z. If this is wrong, then the below answer is wrong and let me know.

* is used for multiplication, not x. This is avoid confusion.

Equation 1 : ( x + x -y = 5 ) becomes ( 2x - y = 5 )

Equation 2 : ( z + y + y = 50 ) becomes ( z + 2y = 50 )

Equation 3 : ( z + z - y = 25 ) becomes ( 2z - y = 25 )

Use Equation 2 to solve for 'z'

z + 2y = 50

-2y -2y

z = 50 - 2y

Plug this value for 'z' into Equation 3

2z - y = 25

2*(50 - 2y) - y = 25

100 - 4y - y = 25

100 - 5y = 25

-100 -100

-5y = -75

divide both sides by -5

y = 15

plug this y into Equation 2

z + 2y = 50

z + 2*15 = 50

z + 30 = 50

-30 -30

z = 20

Plug y into Equation 1

2x - y = 5

2x - 15 = 5

+15 +15

2x = 20

divide both sides by 2

x = 10

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Solve for the '?'

For the last equation you gave :

y + z + x = ? = 15 + 20 + 10 = 45

for the other equation you gave :

( x + z + ? = 40 ) becomes ( ? = 40 - x - z )

? = 40 - 10 - 20 = 10

It is not possible for the 2 '?'s to be the same...

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Always check your work!!

Equation 1 : 2x - y = 2*10 - 15 = 20 - 15 = 5 CORRECT!

Equation 2 : z + 2y = 20 + 2*15 = 20 + 30 = 50 CORRECT!

Equation 3 : 2z - y = 2*20 - 15 = 40 - 15 = 25 CORRECT!

Answer 3

combina variables similares para ecuaciones # uno, dos y cuatro:

2x - y = 5

2y + z = 50

2z - y = 25

En forte, multiplication ecuacione # tres para dos y entonces addicion a ecuacione #dos:

2y + z = 50

2(2z - y) = 25*2

2y + z = 50

4z - 2y = 50

5z = 100 <---- 2y cancelar

z = 20 <-- divido para 5

2y + 20 = 50 <--- substitutos en ecuacion #2

2y = 50 - 20

2y = 30

y = 15

Los substitutos en ecuacion #1

2x - 15 = 5

2x = 5 + 15

2x = 20

x = 10

soluciones

x=10, y=15, z=20

verificacion:

x +x-y = 10 + 10 - 15 = 20 - 15 = 5 <--- si

z + y + y = 20 + 15 + 15 = 35 + 15 = 50 <-- si

z + z - y = 20 + 20 - 15 = 40 - 15 = 25 <--- si

x + z + ? = 40

10 + 20 + ? = 40

30 + ? = 40

? = 10 = x

x + y + z = 10 +15 + 20 = 25 + 20 = 45

finito.

es facil ora????