Plug a few points into the original equations and check to show that the system of equations has one solution. Explain your work.
4x+22=8x+10
and
3x+18=5x+8
To show that the system of equations has one solution, we need to find values of x that satisfy both equations.
For the first equation, 4x + 22 = 8x + 10, let's choose an arbitrary value of x and plug it in:
Let x = 2:
4(2) + 22 = 8(2) + 10
8 + 22 = 16 + 10
30 = 26
As 30 does not equal 26, this value of x does not satisfy the first equation.
For the second equation, 3x + 18 = 5x + 8, let's similarly choose an arbitrary value of x and plug it in:
Let x = 1:
3(1) + 18 = 5(1) + 8
3 + 18 = 5 + 8
21 = 13
Again, 21 does not equal 13, so this value of x does not satisfy the second equation.
Therefore, we cannot find values of x that satisfy both equations simultaneously, indicating that the system of equations does not have one solution.
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.
Sure! Here are three systems of linear equations that correspond to real-world scenarios:
1. System with Infinite Solutions:
Let's consider a scenario where you're at a candy store buying chocolate bars and gummy bears. Let's say that each chocolate bar costs $2 and each gummy bear costs $1. You bought a total of 10 items and spent $15 in total.
The system of equations representing this scenario would be:
2x + y = 15 (equation 1, representing the total cost)
x + y = 10 (equation 2, representing the total number of items)
To solve this system using elimination method, we'll multiply equation 2 by -2 and add it to equation 1:
2x + y = 15
-2(x + y) = -2(10)
This simplifies to:
2x + y = 15
-2x - 2y = -20
Adding the two equations, we get:
-1y = -5
Dividing both sides by -1, we find that y = 5. Substituting y = 5 into equation 2, we find:
x + 5 = 10
x = 5
So, the solution to this system is x = 5 and y = 5. This means that you bought 5 chocolate bars and 5 gummy bears, and this is just one possible solution. If we multiply both sides of equation 2 by any non-zero constant, we would still get the same solution. Therefore, this system has an infinite number of solutions.
2. System with No Solution:
Let's consider a scenario where you're hosting a lemonade stand and selling two sizes of cups – small and large. Let's say you sold a total of 20 cups and earned $50 in total. Each small cup is priced at $2 and each large cup is priced at $3.
The system of equations representing this scenario would be:
2x + 3y = 50 (equation 1, representing the total earnings)
x + y = 20 (equation 2, representing the total number of cups sold)
To solve this system using elimination method, we'll multiply equation 2 by -2 and add it to equation 1:
2x + 3y = 50
-2(x + y) = -2(20)
This simplifies to:
2x + 3y = 50
-2x - 2y = -40
Adding the two equations, we get:
y = 10
Substituting y = 10 into equation 2:
x + 10 = 20
x = 10
But if we substitute these values into equation 1, we find:
2(10) + 3(10) = 50
20 + 30 = 50
50 = 50
Since this equation is true, this system has a unique solution, x = 10 and y = 10.
3. System with One Solution:
Consider a scenario where you have two bank accounts: a checking account and a savings account. The total amount of money you have is $500. In your checking account, you have three times as much as you have in your savings account. The amount in the savings account is unknown.
The system of equations representing this scenario would be:
x + y = 500 (equation 1, representing the total amount of money)
x = 3y (equation 2, representing the relationship between the checking and savings accounts)
Let's solve this system using substitution method. Rearrange equation 2 to solve for x:
x = 3y
Now substitute this expression for x into equation 1:
3y + y = 500
4y = 500
y = 125
Substituting y = 125 into equation 2:
x = 3(125)
x = 375
Thus, the solution to this system is x = 375 and y = 125. This indicates that you have $375 in your checking account and $125 in your savings account.
To check if the system of equations has one solution, we need to plug a few points into the original equations and see if they hold true for both equations simultaneously.
Let's choose x = 0 as our point of comparison.
For the first equation:
4x + 22 = 8x + 10
4(0) + 22 = 8(0) + 10
22 = 10
However, the equation is not true when x = 0.
Now, let's try the second equation:
3x + 18 = 5x + 8
3(0) + 18 = 5(0) + 8
18 = 8
Again, the equation is not true when x = 0.
Since none of our chosen points satisfy both equations, it means that the system of equations does not have one solution.
To show that the system of equations has one solution, we need to plug a few points into the original equations and demonstrate that they satisfy both equations simultaneously.
Let's start by plugging in x = 1 into the equations:
For the first equation, 4x + 22 = 8x + 10 becomes 4(1) + 22 = 8(1) + 10, which simplifies to 4 + 22 = 8 + 10, or 26 = 18.
For the second equation, 3x + 18 = 5x + 8 becomes 3(1) + 18 = 5(1) + 8, which simplifies to 3 + 18 = 5 + 8, or 21 = 13.
As we can see, these values of x do not satisfy both equations simultaneously, since neither equation is true when x = 1. Therefore, x = 1 is not a solution for the system of equations.
Now let's try plugging in x = 2:
For the first equation, 4x + 22 = 8x + 10 becomes 4(2) + 22 = 8(2) + 10, which simplifies to 8 + 22 = 16 + 10, or 30 = 26.
For the second equation, 3x + 18 = 5x + 8 becomes 3(2) + 18 = 5(2) + 8, which simplifies to 6 + 18 = 10 + 8, or 24 = 18.
Again, these values of x do not satisfy both equations simultaneously. Therefore, x = 2 is not a solution for the system of equations.
Let's try one more value of x. Plugging in x = 4:
For the first equation, 4x + 22 = 8x + 10 becomes 4(4) + 22 = 8(4) + 10, which simplifies to 16 + 22 = 32 + 10, or 38 = 42.
For the second equation, 3x + 18 = 5x + 8 becomes 3(4) + 18 = 5(4) + 8, which simplifies to 12 + 18 = 20 + 8, or 30 = 28.
Once again, these values of x do not satisfy both equations simultaneously. Therefore, x = 4 is not a solution for the system of equations.
Based on our attempts, we can conclude that the system of equations does not have a single solution. The given equations are inconsistent and do not intersect at any point on the x-y plane.