5x+5=5x+4 2x+3=2x+7
the original equations and check to show that they can have no
solution. Explain your work.
For the first equation, we can subtract 5x from both sides to get:
5x + 5 - 5x = 5x + 4 - 5x
5 = 4
This statement is not true, so there is no solution for this equation.
For the second equation, we can subtract 2x from both sides to get:
2x + 3 - 2x = 2x + 7 - 2x
3 = 7
This statement is also not true, so there is no solution for this equation as well.
In both cases, the equations start with the same value on both sides, but lead to contradictory statements. This means that there is no value of x that would make the equations true simultaneously. Therefore, these equations have no solution.
Plug a few points into the original equations and check to show that the system of
equations has one solution. Explain your work
3x+18=5x+8 4x+22=8x+10
For the first equation, let's plug in a few points to see if we can find a solution:
Let's try x = 1:
3(1) + 18 = 5(1) + 8
3 + 18 = 5 + 8
21 = 13
The equation is not true, so x = 1 is not a solution.
Now let's try x = 2:
3(2) + 18 = 5(2) + 8
6 + 18 = 10 + 8
24 = 18
Again, the equation is not true for x = 2.
We can continue this process and try different values of x, but we will not find a single value that satisfies the equation. So, the first equation does not have a solution.
For the second equation, let's do the same process:
Let's try x = 1:
4(1) + 22 = 8(1) + 10
4 + 22 = 8 + 10
26 = 18
The equation is not true, so x = 1 is not a solution.
Now let's try x = 2:
4(2) + 22 = 8(2) + 10
8 + 22 = 16 + 10
30 = 26
Again, the equation is not true for x = 2.
Similarly, we can continue trying different values of x, but we will not find a single value that satisfies the equation. So, the second equation does not have a solution either.
In both cases, none of the values we tried satisfied the equations, which means that there is no single value of x that makes the equations true. Therefore, these systems of equations have no solutions.
Writing Linear Equations
Now try writing and solving your own systems of linear equations: one with infinite
solutions, one with no solution, and one with one solution. Write a real-world scenario
that goes with at least one system of equations. Show your work
System 1: Infinite Solutions
Real-world scenario: The cost of a burger is $4 and the cost of a drink is $2. You bought a burger and a drink for a total of $6. Write a system of equations to represent the situation.
Let x represent the number of burgers and y represent the number of drinks.
The cost equation: 4x + 2y = 6
The quantity equation: x + y = 2
To solve this system, we can use the substitution method.
First, solve the quantity equation for x:
x = 2 - y
Next, substitute this expression for x in the cost equation:
4(2 - y) + 2y = 6
Simplify:
8 - 4y + 2y = 6
-2y + 8 = 6
-2y = -2
y = 1
Substitute y = 1 into the quantity equation to find x:
x + 1 = 2
x = 1
So, the solution to this system is x = 1 and y = 1.
System 2: No Solution
Real-world scenario: A farmer has two fields. In field A, he grows wheat, and in field B, he grows corn. The total harvest of wheat and corn is 1000 bushels. The ratio of wheat to corn in field A is 3:2, and the ratio in field B is 2:3. Write a system of equations to represent the situation.
Let x represent the number of bushels of wheat in field A and y represent the number of bushels of corn in field B.
The ratio equation for field A: (3/5)x = (2/5)(1000 - y)
The ratio equation for field B: (2/5)(1000 - y) = (3/5)y
To solve this system, we can multiply both sides of the second equation by 5 to eliminate fractions.
First, distribute and solve for x in the ratio equation for field A:
(3/5)x = (2/5)(1000 - y)
3x = 2(1000 - y)
3x = 2000 - 2y
Now, distribute the 2/5 to simplify the equation for field B:
(2/5)(1000 - y) = (3/5)y
(2/5)1000 - (2/5)y = (3/5)y
400 - (2/5)y = (3/5)y
Combine like terms:
(3/5)y + (2/5)y = 400
Simplify:
(5/5)y = 400
y = 400
Now substitute y = 400 into the equation for field A to find x:
3x = 2000 - 2(400)
3x = 2000 - 800
3x = 1200
x = 400
However, this solution is incorrect because the ratio equation for field A becomes undefined when y = 400.
Therefore, there is no solution to this system of equations.
System 3: One Solution
Real-world scenario: A phone plan offers unlimited texts for a monthly fee of $10 and unlimited data for a monthly fee of $20. You want to find the number of texts and amount of data you used in a month, given that your total bill is $30. Write a system of equations to represent the situation.
Let x represent the number of texts used and y represent the amount of data used.
The cost equation: 10 + 20 = 30
The quantity equation: x + y = 30
To solve this system, we can use the substitution method.
Solve the quantity equation for x:
x = 30 - y
Substitute this expression for x in the cost equation:
10 + 20 = 30 - y
30 = 30 - y
Simplify:
0 = -y
This equation tells us that y must be 0.
Now substitute y = 0 into the quantity equation to find x:
x + 0 = 30
x = 30
So, the solution to this system is x = 30 (texts used) and y = 0 (data used).
To check if these equations have no solutions, we need to simplify and compare the resulting equations. Let's work through them step by step:
1) 5x + 5 = 5x + 4
To solve this equation, we need to isolate the variable x.
Subtracting 5x from both sides, we get:
5x - 5x + 5 = 5x - 5x + 4
Simplifying, we get:
5 = 4
This is not possible since 5 is not equal to 4. Therefore, this equation has no solution.
2) 2x + 3 = 2x + 7
Similarly, we need to isolate the variable x in this equation.
Subtracting 2x from both sides, we get:
2x - 2x + 3 = 2x - 2x + 7
Simplifying, we get:
3 = 7
Again, this equation is not true since 3 is not equal to 7. Therefore, this equation has no solution.
In summary, by simplifying both equations, we found that in both cases, an equality that is not possible arises. Thus, both equations have no solution.
To determine if the given equations have no solution, we need to compare the coefficients of x and the constants on both sides of the equations. If the coefficients and constants are equal, the equations have infinitely many solutions. However, if the coefficients are equal but the constants are not, or if the coefficients are not equal, then the equations have no solution.
Let's analyze the given equations:
1) 5x + 5 = 5x + 4
Here, both equations have the same coefficient of x, i.e., 5, which means we need to compare the constants. By comparing the constants, we see that the left side has a constant of 5, while the right side has a constant of 4. Since the constants are not equal, this equation has no solution.
2) 2x + 3 = 2x + 7
Similarly, for this equation, both sides have the same coefficient of x, 2, so we compare the constants. In this case, the left side has a constant of 3, while the right side has a constant of 7. Since the constants are not equal, this equation also has no solution.
In both cases, the coefficients of x are the same, but the constants are not equal. Thus, there is no value of x that satisfies these equations simultaneously, which confirms that they have no solution.