The required solution to the given system of equations is (3, 4, 1)
a) Given the systems of equations
2x +y +z = 11 .............................. 1
3x + 5y + 5z = 34 .......................2
5x + 6y + 3z = 42 ..........................3
Reduce the given equation to two equations with two unknown:
Multiply equation 1 by 5 and 2 by 1 to have;
2x +y +z = 11 .............................. 1 × 5
3x + 5y + 5z = 34 .......................2 ×1
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10x +5y +5z = 55 .............................. 4
3x + 5y + 5z = 34 .......................5
Subtract equation 4 from 5:
10x - 3x = 55 - 34
7x = 21
x = 21/7
x = 3
Substitue x = 3 into equation 2 and 3 to have:
3(3) + 5y + 5z = 34 .......................6
5(3) + 6y + 3z = 42........................7
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5y + 5z = 25
6y + 3z = 27
The resulting equations can be reduced to;
y + z = 5
2y + z = 9
Subtracting both expressions
y - 2y = 5 - 9
-y = - 4
y = 4
Substitute x = 3 and y = 4 into equation 1;
2x +y +z = 11
2(3) + 4 + z = 11
6 + 4 + z = 11
z = 11-10
z = 1
Hence the required solution to the given system of equations is (3, 4, 1)
b) Check:
Substitute the solutions into equation 1;
2x +y +z = 11
2(3) +4 + 1
6 + 4 + 1 = 11
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