Question

What is the value of y in this linear system? 2x+3y+7z=13, x-2y=3, y+4z=1

Answer 1

To solve this linear system for the value of y, we can use the method of substitution or elimination. Let's use the elimination method. Here's how you can do it:

First, we can solve the second equation (x - 2y = 3) for x:

x = 3 + 2y

Now, we can substitute this expression for x into the first equation (2x + 3y + 7z = 13):

2(3 + 2y) + 3y + 7z = 13

Simplify the equation:

6 + 4y + 3y + 7z = 13

Combine like terms:

7y + 7z = 13 - 6

7y + 7z = 7

Now, we can solve the third equation (y + 4z = 1) for y:

y = 1 - 4z

Substitute this expression for y into the equation we derived:

7(1 - 4z) + 7z = 7

Now, simplify and solve for z:

7 - 28z + 7z = 7

Combine like terms:

-21z = 0

Now, solve for z by dividing both sides by -21:

z = 0

Now that we have the value of z, we can find the value of y using the third equation (y + 4z = 1):

y + 4(0) = 1

y = 1

So, the value of y in the linear system is y = 1.

Answer 2

y = 1

Explanation:

2x + 3y + 7z = 13

x - 2y = 3

y + 4z = 1

Solve the third equation for z.

4z = 1 - y

z = 1/4(1 - y)

Substitute 1/4(1 - y) for z in the first equation.

2x + 3y + 7(1/4)(1 - y) = 13

Multiply both sides by 4.

8x + 12y + 7 - 7y = 52

8x + 5y = 45

8x + 5y = 45

x - 2y = 3

8x + 5y = 45

-8x + 16y = -24

21y = 21

y = 1