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

Skip to content# 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,

Mathematics ##
Comments (5) on “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,”

### Leave a Reply Cancel reply

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

3432532

step-by-step explanation:

search in google or doubtnut for better answers

what are the statments

step-by-step explanation:

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

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