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# help

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x + x + y = 14

L + L + z = 20

x + y + z = 12

x + z + L = 18

Find x+ z.

Dec 26, 2019

#1
+1

2x+y=14 (1)

L+L+z=20 (2)

x+y+z=12 (3)

x+z+L=18 (4)

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

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

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

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

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

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

Dec 26, 2019
edited by Guest  Dec 26, 2019
#2
0

Confirmation from Wolfram:

https://www.wolframalpha.com/input/?i=systems+of+equations+calculator&assumption=%22FSelect%22+-%3E+%7B%7B%22SolveSystemOf4EquationsCalculator%22%7D%7D&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation1%22%7D+-%3E%222x%2By%3D14%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation2%22%7D+-%3E%22L%2BL%2Bz%3D20%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation3%22%7D+-%3E%22x%2By%2Bz%3D12%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation4%22%7D+-%3E%22x%2BL%2Bz%3D18%22

Guest Dec 26, 2019