Solve the system by using a table.
1-0.6x - 0.6y =-0.6
10.5x-0.3▶=-27
To solve the system using a table, we'll first create a table and set up the equations:
Equation 1: 1-0.6x - 0.6y = -0.6
Equation 2: 10.5x-0.3y = -27
We'll start by selecting values for x and finding the corresponding y values for each equation. Let's choose x = 0, 1, and 2:
When x = 0:
Equation 1: 1-0.6(0) - 0.6y = -0.6
Simplifying, we get: 1 - 0 - 0.6y = -0.6
Simplifying further, we get: 1 - 0.6y = -0.6
Rearranging the equation, we get: 1 + 0.6y = -0.6
Subtracting 1 from both sides, we get: 0.6y = -1.6
Dividing both sides by 0.6, we get: y = -1.6/0.6 ≈ -2.67
Equation 2: 10.5(0) - 0.3y = -27
Simplifying, we get: -0.3y = -27
Dividing both sides by -0.3, we get: y = -27/-0.3 = 90
So, when x = 0, y ≈ -2.67 and y = 90.
When x = 1:
Equation 1: 1-0.6(1) - 0.6y = -0.6
Simplifying, we get: 1 - 0.6 - 0.6y = -0.6
Simplifying further, we get: 0.4 - 0.6y = -0.6
Rearranging the equation, we get: 0.4 + 0.6y = -0.6
Subtracting 0.4 from both sides, we get: 0.6y = -1
Dividing both sides by 0.6, we get: y = -1/0.6 ≈ -1.67
Equation 2: 10.5(1) - 0.3y = -27
Simplifying, we get: 10.5 - 0.3y = -27
Subtracting 10.5 from both sides, we get: -0.3y = -37.5
Dividing both sides by -0.3, we get: y = -37.5/-0.3 = 125
So, when x = 1, y ≈ -1.67 and y = 125.
When x = 2:
Equation 1: 1-0.6(2) - 0.6y = -0.6
Simplifying, we get: 1 - 1.2 - 0.6y = -0.6
Simplifying further, we get: -0.2 - 0.6y = -0.6
Rearranging the equation, we get: -0.2 + 0.6y = -0.6
Adding 0.2 to both sides, we get: 0.6y = -0.4
Dividing both sides by 0.6, we get: y = -0.4/0.6 ≈ -0.67
Equation 2: 10.5(2) - 0.3y = -27
Simplifying, we get: 21 - 0.3y = -27
Subtracting 21 from both sides, we get: -0.3y = -48
Dividing both sides by -0.3, we get: y = -48/-0.3 = 160
So, when x = 2, y ≈ -0.67 and y = 160.
Based on the values we calculated, the solutions to the system of equations are:
(x, y) ≈ (0, -2.67), (1, -1.67), (2, -0.67)
(x, y) = (0, 90), (1, 125), (2, 160)
To solve the system using a table, we need to create a table with columns for x and y values. Then, we will substitute different values for x and solve for y for each value of x.
Let's start by substituting x = 0 into the first equation and solving for y.
1 - 0.6(0) - 0.6y = -0.6
1 - 0 - 0.6y = -0.6
1 - 0.6y = -0.6
-0.6y = -0.6 - 1
-0.6y = -1.6
y = -1.6 / -0.6
y ≈ 2.67
Now, let's substitute x = 1 into the first equation and solve for y.
1 - 0.6(1) - 0.6y = -0.6
1 - 0.6 - 0.6y = -0.6
0.4 - 0.6y = -0.6
-0.6y = -0.6 - 0.4
-0.6y = -1
y = -1 / -0.6
y ≈ 1.67
We will continue substituting different values for x and solving for y. However, there seems to be an error in the second equation you provided. It is written as "10.5x-0.3▶=-27", and I'm unable to interpret it correctly. Could you please double-check and provide the correct equation?
To solve the system using a table, we need to express both equations in the form "y = mx + b".
The given equations are:
1) 1 - 0.6x - 0.6y = -0.6
2) 10.5x - 0.3y = -27
Let's rearrange equation 1 to get y in terms of x:
1 - 0.6x - 0.6y = -0.6
Subtract 1 from both sides:
-0.6x - 0.6y = -1.2
Divide by -0.6:
x + y = 2
Now, rewrite equation 2 in terms of y:
10.5x - 0.3y = -27
Subtract 10.5x from both sides:
-0.3y = -10.5x - 27
Divide by -0.3:
y = 35x + 90
To create a table, we will assign values to x and then use these values to find corresponding y values.
Let's choose three values for x and calculate the corresponding y values:
When x = 0:
Using equation 1:
x + y = 2
0 + y = 2
y = 2
Using equation 2:
y = 35x + 90
y = 35(0) + 90
y = 90
The first row of the table is (0, 2, 90).
When x = 1:
Using equation 1:
x + y = 2
1 + y = 2
y = 1
Using equation 2:
y = 35x + 90
y = 35(1) + 90
y = 35 + 90
y = 125
The second row of the table is (1, 1, 125).
When x = -1:
Using equation 1:
x + y = 2
-1 + y = 2
y = 3
Using equation 2:
y = 35x + 90
y = 35(-1) + 90
y = -35 + 90
y = 55
The third row of the table is (-1, 3, 55).
Now, we have the following table:
x | y (from equation 1) | y (from equation 2)
-----------------------------------------
0 | 2 | 90
1 | 1 | 125
-1| 3 | 55
By using the table, we can see that the solutions to the system are:
(0, 2)
(1, 1)
(-1, 3)