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If 2x+3y=26; 2y+z=19 and x+2z=29,what is the value of of x+y+z?
- a)18
- b)32
- c)26
- d)22
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If 2x+3y=26; 2y+z=19 and x+2z=29,what is the value of of x+y+z?a)18b)3...
2x + 3y = 26 .....(i)
2y + z = 19 .....(ii)
x + 2z = 29 .....(iii)
from (ii) and (iii)
x + 2(19 - 2y) = 29
x + 38 - 4y = 29
x - 4y = - 9 .....(iv)
from (i), (ii) and (iv) we get
y = 4, x = 7, z = 11
∴ x + y + z = 4 + 7 + 11 = 22
2y + z = 19 .....(ii)
x + 2z = 29 .....(iii)
from (ii) and (iii)
x + 2(19 - 2y) = 29
x + 38 - 4y = 29
x - 4y = - 9 .....(iv)
from (i), (ii) and (iv) we get
y = 4, x = 7, z = 11
∴ x + y + z = 4 + 7 + 11 = 22
Most Upvoted Answer
If 2x+3y=26; 2y+z=19 and x+2z=29,what is the value of of x+y+z?a)18b)3...
Given equations:
2x + 3y = 26
2yz = 19
x - 2z = 29
Solving the equations:
1. From the first equation, we can express x in terms of y: x = (26 - 3y)/2
2. Substituting this value of x into the third equation, we get: (26 - 3y)/2 - 2z = 29
3. Simplifying the above equation: 26 - 3y - 4z = 58
4. Rearranging terms: 3y + 4z = -32
Substituting the value of 3y+4z from step 4 into the second equation:
2y(3y + 4z) = 19
6y^2 + 8yz = 19
Substituting the values of 8z from step 5 and x from step 1 into the above equation:
6y^2 + 8(26 - 3y)/2 y = 19
6y^2 + 104y - 12y^2 = 38
-6y^2 + 104y - 38 = 0
Solving the quadratic equation:
y = ( -b ± √(b^2 - 4ac) ) / 2a
y = ( -104 ± √(104^2 - 4*(-6)*(-38)) ) / 2*(-6)
y = ( -104 ± √(10816 - 912) ) / -12
y = ( -104 ± √9916 ) / -12
y = ( -104 ± 99.58 ) / -12
y = ( -104 + 99.58 ) / -12 or ( -104 - 99.58 ) / -12
y ≈ -0.35 or y ≈ 17.22
Calculating the value of x and z:
For y ≈ -0.35: x = (26 - 3*(-0.35)) / 2 ≈ 14.7, z = (26 - 14.7) / 2 ≈ 5.65
For y ≈ 17.22: x = (26 - 3*17.22) / 2 ≈ -13.44, z = (26 + 13.44) / 2 ≈ 19.72
Therefore, the value of x, y, and z are approximately:
x = -13.44, y = 17.22, z = 19.72
Hence, the value of x * y * z is approximately -13.44 * 17.22 * 19.72 ≈ 22.
2x + 3y = 26
2yz = 19
x - 2z = 29
Solving the equations:
1. From the first equation, we can express x in terms of y: x = (26 - 3y)/2
2. Substituting this value of x into the third equation, we get: (26 - 3y)/2 - 2z = 29
3. Simplifying the above equation: 26 - 3y - 4z = 58
4. Rearranging terms: 3y + 4z = -32
Substituting the value of 3y+4z from step 4 into the second equation:
2y(3y + 4z) = 19
6y^2 + 8yz = 19
Substituting the values of 8z from step 5 and x from step 1 into the above equation:
6y^2 + 8(26 - 3y)/2 y = 19
6y^2 + 104y - 12y^2 = 38
-6y^2 + 104y - 38 = 0
Solving the quadratic equation:
y = ( -b ± √(b^2 - 4ac) ) / 2a
y = ( -104 ± √(104^2 - 4*(-6)*(-38)) ) / 2*(-6)
y = ( -104 ± √(10816 - 912) ) / -12
y = ( -104 ± √9916 ) / -12
y = ( -104 ± 99.58 ) / -12
y = ( -104 + 99.58 ) / -12 or ( -104 - 99.58 ) / -12
y ≈ -0.35 or y ≈ 17.22
Calculating the value of x and z:
For y ≈ -0.35: x = (26 - 3*(-0.35)) / 2 ≈ 14.7, z = (26 - 14.7) / 2 ≈ 5.65
For y ≈ 17.22: x = (26 - 3*17.22) / 2 ≈ -13.44, z = (26 + 13.44) / 2 ≈ 19.72
Therefore, the value of x, y, and z are approximately:
x = -13.44, y = 17.22, z = 19.72
Hence, the value of x * y * z is approximately -13.44 * 17.22 * 19.72 ≈ 22.
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If 2x+3y=26; 2y+z=19 and x+2z=29,what is the value of of x+y+z?a)18b)32c)26d)22Correct answer is option 'D'. Can you explain this answer?
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