The given equations are:
1) 2x + 3 = 2x + 7
2) 3x + 18 = 5x + 8
Let's solve each equation one by one:
1) 2x + 3 = 2x + 7
Subtracting 2x from both sides, we get:
3 = 7
Since 3 is not equal to 7, this equation has no solution.
2) 3x + 18 = 5x + 8
Subtracting 3x from both sides, we get:
18 = 2x + 8
Subtracting 8 from both sides, we get:
10 = 2x
Dividing both sides by 2, we get:
5 = x
So, the solution for the second equation is x = 5.
To summarize, the system of equations has no solution for the first equation and x = 5 for the second equation.
Consider the two equations
2x+3=2x+7
3x+18=5x+8
you put a square around, which form a system with no solution. Solve the equations.
9 answers
my bad the first equation was supposed to be
4x+22=8x+10
4x+22=8x+10
No problem! Let's solve the revised system of equations:
1) 4x + 22 = 8x + 10
To solve for x, we can start by simplifying the equation:
Subtracting 4x from both sides, we get:
22 = 4x + 10
Subtracting 10 from both sides, we get:
12 = 4x
Dividing both sides by 4, we get:
x = 3
So, the solution for the first equation is x = 3.
2) 3x + 18 = 5x + 8
Subtracting 3x from both sides, we get:
18 = 2x + 8
Subtracting 8 from both sides, we get:
10 = 2x
Dividing both sides by 2, we get:
5 = x
So, the solution for the second equation is x = 5.
To summarize, the solutions for the revised system of equations are x = 3 and x = 5.
1) 4x + 22 = 8x + 10
To solve for x, we can start by simplifying the equation:
Subtracting 4x from both sides, we get:
22 = 4x + 10
Subtracting 10 from both sides, we get:
12 = 4x
Dividing both sides by 4, we get:
x = 3
So, the solution for the first equation is x = 3.
2) 3x + 18 = 5x + 8
Subtracting 3x from both sides, we get:
18 = 2x + 8
Subtracting 8 from both sides, we get:
10 = 2x
Dividing both sides by 2, we get:
5 = x
So, the solution for the second equation is x = 5.
To summarize, the solutions for the revised system of equations are x = 3 and x = 5.
what do you notice about these equations
Upon examining the equations, we can notice that the coefficients of x on both sides of the equation are the same for both equations. This means that if we subtract 2x from both sides, the resulting equations will be the same for both equations.
For the first equation, subtracting 2x from both sides gives:
3 = 7
This results in a contradiction since 3 is not equal to 7. Thus, the first equation is inconsistent and has no solution.
For the second equation, subtracting 2x from both sides gives:
18 = 8
Again, this results in a contradiction since 18 is not equal to 8. Thus, the second equation is also inconsistent and has no solution.
In summary, both equations are inconsistent and do not have a valid solution.
For the first equation, subtracting 2x from both sides gives:
3 = 7
This results in a contradiction since 3 is not equal to 7. Thus, the first equation is inconsistent and has no solution.
For the second equation, subtracting 2x from both sides gives:
18 = 8
Again, this results in a contradiction since 18 is not equal to 8. Thus, the second equation is also inconsistent and has no solution.
In summary, both equations are inconsistent and do not have a valid solution.
what does it mean to have one solution?
Having one solution means that the system of equations has a unique solution. It means that there is a specific value for the variable(s) that satisfies both equations simultaneously.
In a system of linear equations with two variables (x and y), having one solution means that there is only one point on the coordinate plane where the graphs of the two equations intersect. This point represents the values of x and y that satisfy both equations.
Graphically, this intersection point represents the solution to the system, and it is the only point where the two lines corresponding to the equations intersect.
Algebraically, it means that when solving the system of equations, you can obtain a unique value for each variable that makes both equations true when substituted into the equations.
In a system of linear equations with two variables (x and y), having one solution means that there is only one point on the coordinate plane where the graphs of the two equations intersect. This point represents the values of x and y that satisfy both equations.
Graphically, this intersection point represents the solution to the system, and it is the only point where the two lines corresponding to the equations intersect.
Algebraically, it means that when solving the system of equations, you can obtain a unique value for each variable that makes both equations true when substituted into the 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.
Sure! Here are three systems of linear equations, each with a different number of solutions:
1) Infinite Solutions:
Equation 1: 2x + 4y = 8
Equation 2: 4x + 8y = 16
This system of equations represents two parallel lines that have the same slope and intersect at infinitely many points. Any values of x and y that satisfy one equation will also satisfy the other equation. These equations could represent two similar recipes that yield the same result when doubled or tripled.
To solve this system, we can simplify Equation 1:
2x + 4y = 8
Divide by 2:
x + 2y = 4
Since Equation 2 is a multiple of Equation 1, we can see that these equations are equivalent. They have the same solution set: infinite solutions.
2) No Solution:
Equation 1: 3x + 2y = 10
Equation 2: 6x + 4y = 20
This system of equations represents two parallel lines that never intersect. They have the same slope but different y-intercepts. These equations could represent two paths or journeys that will never cross.
To solve this system, we can simplify Equation 1:
3x + 2y = 10
Divide by 3:
x + (2/3)y = 10/3
If we simplify Equation 2:
6x + 4y = 20
Divide by 6:
x + (2/3)y = 10/3
We can see that both equations are equivalent. This implies that the system has no solution.
3) One Solution:
Equation 1: 2x + y = 5
Equation 2: 3x - y = 7
This system of equations can be represented as two lines that intersect at a single point. It could describe a situation where the cost (y) of a product is based on the number of units (x) purchased and another line representing a budget constraint.
To solve this system, we can use the method of elimination:
Multiply Equation 1 by 3 and Equation 2 by 2 to get:
6x + 3y = 15
6x - 2y = 14
Subtracting Equation 2 from Equation 1, we have:
6x + 3y - (6x - 2y) = 15 - 14
5y = 1
y = 1/5
Plugging the value of y back into Equation 1:
2x + (1/5) = 5
2x + 1 = 25/5
2x + 1 = 5
2x = 4
x = 2
The solution to this system is x = 2 and y = 1/5.
I hope this helps! Let me know if you have any more questions or clarifications.
1) Infinite Solutions:
Equation 1: 2x + 4y = 8
Equation 2: 4x + 8y = 16
This system of equations represents two parallel lines that have the same slope and intersect at infinitely many points. Any values of x and y that satisfy one equation will also satisfy the other equation. These equations could represent two similar recipes that yield the same result when doubled or tripled.
To solve this system, we can simplify Equation 1:
2x + 4y = 8
Divide by 2:
x + 2y = 4
Since Equation 2 is a multiple of Equation 1, we can see that these equations are equivalent. They have the same solution set: infinite solutions.
2) No Solution:
Equation 1: 3x + 2y = 10
Equation 2: 6x + 4y = 20
This system of equations represents two parallel lines that never intersect. They have the same slope but different y-intercepts. These equations could represent two paths or journeys that will never cross.
To solve this system, we can simplify Equation 1:
3x + 2y = 10
Divide by 3:
x + (2/3)y = 10/3
If we simplify Equation 2:
6x + 4y = 20
Divide by 6:
x + (2/3)y = 10/3
We can see that both equations are equivalent. This implies that the system has no solution.
3) One Solution:
Equation 1: 2x + y = 5
Equation 2: 3x - y = 7
This system of equations can be represented as two lines that intersect at a single point. It could describe a situation where the cost (y) of a product is based on the number of units (x) purchased and another line representing a budget constraint.
To solve this system, we can use the method of elimination:
Multiply Equation 1 by 3 and Equation 2 by 2 to get:
6x + 3y = 15
6x - 2y = 14
Subtracting Equation 2 from Equation 1, we have:
6x + 3y - (6x - 2y) = 15 - 14
5y = 1
y = 1/5
Plugging the value of y back into Equation 1:
2x + (1/5) = 5
2x + 1 = 25/5
2x + 1 = 5
2x = 4
x = 2
The solution to this system is x = 2 and y = 1/5.
I hope this helps! Let me know if you have any more questions or clarifications.
5 answers