#1**+1 **

2x+y=14 (1)

L+L+z=20 (2)

x+y+z=12 (3)

x+z+L=18 (4)

\(0L+2x+y+0z=14\)

\(2L+0x+0y+z=20\)

\(0L+x+y+z=12\)

\(L+x+0y+z=18\)

Add the 4 equations

\(3L+4x+2y+3z=64\)

3(L+z)+2(2x+y)=64 I factorising by grouping.

We know from (1) that 2x+y=14

so

3(L+z)+2(14)=64

3(L+z)+28=64 I subtract 28

3(L+z)=36 I divide by 3

L+z=12 (5)

So we can find the value of L from (2)

L+L+z=20

we know that L+z=12 from (5)

L+12=20

L=8

Well also from (2) since we knew that L=8 we can now z

8+8+z=20

z=20-16

z=4

from (4) we know that x+z+L=18

so

x+4+8=18

x=18-12

x=6

The question may end here as you asked "Find x+z" we know x and we know z so x+z=6+4=10 , the answer is 10.

However, let's contuine to get the value of y.

from (1) we know that x+x+y=14

we know that x=6

thus,

6+6+y=14

y=14-12

y=2

So,

L=8

x=6

y=2

z=4

Guest Dec 26, 2019

edited by
Guest
Dec 26, 2019

#2**0 **

Confirmation from Wolfram:

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Guest Dec 26, 2019