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What is the solution to the following system? x+2y+=9;x-y+3z=13;2z=10
x = 10, y = 2, z = 5
x = –4, y = 2, z = 5
x = 0, y = 2, z = 5
x = 4, y = 6, z = 5

Respuesta :

mathstudent55
x + 2y + z = 9
x - y + 3z = 13
2z = 10

Solve the 3rd equation for z.
z = 5

Now substitute z = 5 into both the first and seconds equations.


x + 2y + 5 = 9
x - y + 3(5) = 13

x + 2y = 4
x - y = -2

Subtract the first equation from the second equation above.

-3y = -6

y = 2

Now substitute z = 5 and y = -2 into the first original equation, and solve for x.

x + 2(2) + 5 = 9

x + 9 = 9

x = 0

Answer: x = 0; y = 2; z = 5
2z=10
z=10/2=5 so that is easy

substitute z=5 in x-y+3z=13
x-y+3*5=13
x-y=-2

subtract x-y=-2 from x+2y+(?)=9
3y+(?)=11

if (?) is nothing, then y=11/3 or 3.67 and x=5/3 or 1.67

but if (?) is 1z, then 3y+5=11
3y=6
y=2

x=-2+y=-2+2=0

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