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lukranit [14]
4 years ago
10

X+y+z=1

Mathematics
1 answer:
andrezito [222]4 years ago
4 0

Answer: x = 0

y = 2

z = -1

Step-by-step explanation:

The system of equations are

x+y+z=1 - - - - - - - - - - 1

-2x+4y+6z=2 - - - - - - - - - 2

-x+3y-5z=11 - - - - - - - - - 3

Step 1

We would eliminate x by adding equation 1 to equation 3. It becomes

4y -4z = 12 - - - - - - - - - 4

Step 2

We would multiply equation 1 by 2. It becomes

2x + 2y + 2z = 2 - - - - - - - - - 5

We would add equation 2 and equation 5. It becomes

6y + 8z = 4 - - - - - - - - - 6

Step 3

We would multiply equation 4 by 6 and equation 6 by 4. It becomes

24y - 24z = 72 - - - - - - - - 7

24y + 32z = 16 - - - - - - - - 8

We would subtract equation 8 from equation 7. It becomes

-56z = 56

z = -56/56 = -1

Substituting z = -1 into 7, it becomes

24y - 24×-1 = 72

24y + 24 = 72

24y = 72 - 24 = 48

y = 48/24 = 2

Substituting y = 2 and z = -1 into equation 1, it becomes

x + 2 - 1 = 1

x = 1 - 1 = 0

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PLEASE SOLVE AND CHECK. SHOW COMPLETE SOLUTION
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<u>Solution</u><u>:</u>

\sqrt{4x + 13}  = x + 2

  • First square both sides.

=  > ( \sqrt{4x + 13} ) ^{2}  = (x + 2) ^{2}

  • Now, square root and square gets cancel out in the LHS. And in the RHS, apply the identity: (a + b)² = a² + 2ab + b².

=  > 4x + 13 =  {(x)}^{2}  + 2 \times x \times 2 + (2) ^{2} \\  =  > 4x + 13 =  {x}^{2}   + 4x + 4

  • Now, transpose 4x and 4 to LHS.

=  > 4x - 4x + 13 - 4 =  {x}^{2}  \\

  • Now, do the addition and subtraction.

=  >  {x}^{2}  = 9 \\  =  >  x =  \sqrt{9}  \\  =  > x = ±3

<u>Answer</u><u>:</u>

<u>x </u><u>=</u><u> </u><u>±</u><u> </u><u>3</u>

Hope you could understand.

If you have any query, feel free to ask.

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