Answer:
Infinite number of solutions, (x, 8-5x)
Explanation:
This is a "System of Equations" type problem. What it's asking is for you to find if there are no solutions, 1 solution, or infinitely many.
I'll explain what those mean first.
Here's an example of a system with no solutions:
x+y=2
x+y=3
There are no solutions because there are no values that you could assign to x and y that would satisfy both equations. 2 can never equal 3.
Here's an example of a system with one solution:
2x+y=7
x+y=5
There is one solution because once we solve for x and y, the only possible solution we can get is x=2 and y=3, so the one solution is (2,3). I will explain the methods to find the solution in a bit.
Here's an example of a system with infinitely many solutions:
x+y=2
2x+2y=4
There are infinitely many solutions because these two equations lie on the same line. Factor out 2 in the second equation 2x+2y=4 and you get x+y=2, which is the same as the first equation. This means whatever answers works with one equation works for the other. With x+y=2 there are an infinite amount of values for x and y that could work, like (1,1), (0.5, 1.5), (0,2), etc.
For more help search for this video on a video sharing platform that I may or may not be able to say the name of: "One Solution, No Solution, or Infinitely Many Solutions - Consistent & Inconsistent Systems"
When trying to find the solutions to the system there are usually two methods: Substitution, and Elimination.
In Substitution, you would take one equation, get one of the variables in terms of the other variable (such as x in terms of y), then plug it into the other equation to solve for the value of the other variable, and then take that value and plug it back into your original expression.
Example:
x+y=5
2x+y=7
From the first equation, I can say x=5-y. Then I plug my expression of x into the second equation.
2(5-y)+y=7
10-2y+y=7
y=3
Knowing y is 3, we can plug it back into our x expression, to find that x=5-3=2. So the answer is (2, 3).
In Canceling, you take multiples of equations and add them to the other equation in an attempt to cancel a variable so you can solve for the other.
Example:
x+y=6
4x+y=18
I can multiply the entire first equation by -1, leaving me with -x-y=-6. I'm doing this because I see a positive y in the second equation, and if I multiply the first equation by -1, I will have a -y, which will cancel out with the +y when I add the two equations.
If I add the equations now, I get
(-x-y)+(4x+y)=(-6)+18
-x-y+4x+y=12
3x=12
x=4
Then I can take x=4, plug it into the first equation, and get 4+y=6, and find y=2.
For more on these two methods search for: "Solving Systems of Equations By Elimination & Substitution With 2 Variables"
Now, to actually answer your question (sorry if this explanation was long-winded; if you want, just watch those two videos and they should do a good job. The Organic Chemistry Tutor is a great teacher.)
You have
5x+y=8
-10x-2y=-16
Immediately, I notice that the 2nd equation is a multiple of the 1st.
If you multiply the 1st equation 5x+y=8 by -2, you get the 2nd equation.
Rule of thumb: Whenever you have a system of two equations that are multiples of each other, there are infinitely many solutions. Because there are infinitely many solutions, we can't give a single answer, but instead, we can do what the question suggests and give the relationship between x and y in coordinate format. We can do this by simply taking one of the equations and finding y in terms of x.
In the first equation 5x+y=8, we can subtract 5x from both sides and get y=8-5x.
So the final answer would be The system has Infinitely many solutions, and the solution is (x, 8-5x).