What is x+y+z=63,2-4y=1,2y+32=7 ?

The solution to x+y+z=63,2-4y=1,2y+32=7 is No Solution

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x+y+z=63,2-4y=1,2y+32=7

x + y + z = 63, 2 - 4 y = 1, 2 y + 32 = 7

No Solution

Solution steps

x + y + z = 63 2 - 4 y = 1 2 y + 32 = 7

Isolate

y

for

2 - 4 y = 1 : y = \frac{1}{4}

Substitute

y = \frac{1}{4}

[2 \cdot \frac{1}{4} + 32 = 7 x + \frac{1}{4} + z = 63]

Simplify

[\frac{65}{2} = 7 x + \frac{1}{4} + z = 63]

\frac{65}{2} = 7

is false, therefore the system of equations has no solution

No Solution

a = 5, b = 6

3 x = 2, y + 5 z^{3} y = 4 x + z

4 x - 5 y = 25, 5 x - 6 y = 50

x = 150 + (\frac{y}{2}), y = x + 150

30 x + 44 y = 10, 40 x + 55 y = 13

What is x+y+z=63,2-4y=1,2y+32=7 ?
The solution to x+y+z=63,2-4y=1,2y+32=7 is No Solution