Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions,state this.
3x+7y=77
-2x+7y=42
Subtract equation #2 from equation # 1.
5x = 35
You should be able to take it from here.
I understand where I should what from where plz help me out. It would really be appreciaeted.
3x-(-2x) = 3x+2x = 5x
7y-7y = 0
77-42 = 35
Therefore 5x + 0 = 35 or 5x = 35
To solve the system of equations using the elimination method, we need to eliminate one variable by adding or subtracting the two equations. Let's proceed with the following steps:
Step 1: Multiply the second equation by 3 to make the coefficients of x in both equations opposite each other.
Multiply -2x + 7y = 42 by 3:
3(-2x + 7y) = 3(42)
-6x + 21y = 126
Now the system of equations becomes:
3x + 7y = 77
-6x + 21y = 126
Step 2: Add the equations together. The x-term will be eliminated.
(3x + 7y) + (-6x + 21y) = 77 + 126
-3x + 28y = 203
So, our new equation is:
-3x + 28y = 203
Now, let's determine whether this simplified equation has a solution or not.
In order to determine if a system of equations has a solution or infinitely many solutions, we compare the number of variables with the number of distinct equations involving those variables.
In this case, we have two variables (x and y) and only one distinct equation involving them. Therefore, this system has an infinite number of solutions.
To find one such solution, we can express either x or y in terms of the other variable. Let's solve for y in terms of x:
-3x + 28y = 203
Rearrange the equation to solve for y:
28y = 3x + 203
y = (3x + 203)/28
We can choose any value for x, and then substitute it into the equation to find the corresponding value for y. This will give us one solution of infinitely many possibilities.
For example, let's choose x = 1:
y = (3*1 + 203)/28
y = 7
So, one solution to this system of equations is x = 1 and y = 7. But there are infinitely many other solutions due to the nature of the system.