Question

determine the solution to the linear system of equation from the table

x y1 y2
8 8 5
9 12 10
10 16 15
11 20 20
12 24 25

Answer 1

look at the differences

y1 = 4(x-6)
y2 = 5(x-7)

Answer 2

[AI-generated answer — Step-by-Step Bot]

To determine the solution to the linear system of equations from the given table, we need to find the relationship between x and y1 and between x and y2.

First, let's determine the relationship between x and y1:
Looking at the values of x and y1, we can observe that as x increases by 1, y1 increases by a constant difference of 4. This implies that y1 and x have a linear relationship with a slope of 4.

Next, let's determine the relationship between x and y2:
Examining the values of x and y2, we can see that as x increases by 1, y2 increases by a constant difference of 5. This indicates that y2 and x have a linear relationship with a slope of 5.

So, we can express the equations for y1 and y2 in terms of x as follows:
y1 = 4x
y2 = 5x

Now we have the equations for y1 and y2, we can solve the linear system of equations by finding the common solution for x that will satisfy both equations.

Since y1 = y2, we can set 4x = 5x:
4x - 5x = 0
-x = 0
x = 0

Therefore, the solution for x is 0.

To find the corresponding values of y1 and y2, we can substitute x = 0 into either of the equations:
y1 = 4(0) = 0
y2 = 5(0) = 0

Hence, the solution to the linear system of equations is x = 0, y1 = 0, and y2 = 0.

Answer 3

[AI-generated answer — Explain Bot]

To determine the solution to the linear system of equations from the given table, we can use the method of substitution.

First, let's write down the equations for the system:

Equation 1: y1 = mx + b1
Equation 2: y2 = mx + b2

To solve for the constants m, b1, and b2, we need to use the given table values.

Using the first row of the table:
x = 8, y1 = 8, y2 = 5

Substituting these values into Equation 1:
8 = 8m + b1

Substituting these values into Equation 2:
5 = 8m + b2

Now, we need to repeat this process using the second row of the table and then the third row of the table. Finally, we will have a system of three equations that can be solved simultaneously.

Using the second row of the table:
x = 9, y1 = 12, y2 = 10

Substituting these values into Equation 1:
12 = 9m + b1

Substituting these values into Equation 2:
10 = 9m + b2

Using the third row of the table:
x = 10, y1 = 16, y2 = 15

Substituting these values into Equation 1:
16 = 10m + b1

Substituting these values into Equation 2:
15 = 10m + b2

Now, we have a system of three equations:
1) 8 = 8m + b1
2) 5 = 8m + b2
3) 16 = 10m + b1
4) 15 = 10m + b2

We can solve this system by subtracting Equation 1 from Equation 3 and also subtracting Equation 2 from Equation 4 to eliminate the b1 and b2 terms:

3) 16 = 10m + b1
1) - 8 = -8m - b1
---------------------
5) 8 = 2m

4) 15 = 10m + b2
2) - 5 = -8m - b2
--------------------
6) 10 = 2m + b2

Now, we have a system of two equations:
5) 8 = 2m
6) 10 = 2m + b2

From Equation 5, we can solve for m:
2m = 8
m = 4

Substituting m = 4 into Equation 6, we can solve for b2:
10 = 2(4) + b2
10 = 8 + b2
b2 = 2

Now that we have the values of m and b2, we can substitute them back into Equation 1 to solve for b1:
8 = 8(4) + b1
8 = 32 + b1
b1 = -24

So, the solution to the linear system of equations is:
y1 = 4x - 24
y2 = 4x + 2