Find the number of solutions for this system of equations.
3x + 5y = 20
-2x +5y = 20
How can I do this (and for other systems of equations) without taking all day? Thanks!
In this particular case, the problem tells you that both formulas equal 20. This means that 3x+5y = -2x +5y.
If you put the ys on one side and the xs on the other, the 5ys disappear.
3x + 5y -5y = -2x + 5y -5y
(what you do on one side, has to be done on the other).
That leaves 3x = -2x.
To get rid of the -2x on the right side, you would add 2x to both sides.
This gives
3x + 2x = -2x +2x
which simplifies to
5x = 0
The only number that works for x here is 0.
If you plug zero in for x in either of the original equations, you then get y = 4.
Obviously both equations cannot be true unless x = 0. That means, also, that
5y = 20.
One more step and you can get y.
That didn't take all day.
That is the only one solution.
Actually you were not even asked for what x and y are. You were asked how many solutions there are. Each equation is a straight line on a graph, but they have different slopes. At how many points can two such straight lines intersect?
To find the number of solutions for a system of equations, you can use the method of elimination or substitution. Here's a step-by-step process to solve this system using elimination:
Step 1: Multiply one or both equations by constant(s) to ensure that the coefficients of either x or y will cancel out when added.
In this case, we can see that the coefficients of y are already the same, so we can proceed with the elimination method without multiplying any equations.
Step 2: Add or subtract the equations to eliminate one variable.
To eliminate the x variable, subtract the second equation from the first equation:
(3x + 5y) - (-2x + 5y) = 20 - 20
3x + 5y + 2x - 5y = 0
5x = 0
Simplifying the equation, we find that 5x = 0.
Step 3: Solve for the remaining variable.
Divide both sides of the equation by 5:
5x/5 = 0/5
x = 0
Step 4: Substitute the value of x back into one of the original equations to solve for the remaining variable.
Using the first equation:
3x + 5y = 20
3(0) + 5y = 20
0 + 5y = 20
5y = 20
Step 5: Solve for y.
Divide both sides of the equation by 5:
5y/5 = 20/5
y = 4
Therefore, the solution for the system of equations is x = 0, y = 4.
Now, let's consider the number of solutions in a general sense:
If you have one unique solution, the system is consistent and independent. This means that the graphs of the equations intersect at a single point.
If you have no solution, the system is inconsistent. This means that the graphs of the equations do not intersect and are parallel.
If you have infinitely many solutions, the system is consistent and dependent. This means that the graphs of the equations coincide, and every point on one line is also on the other line.
In this case, we have a consistent and independent system since the equations intersect at a single point (0, 4). So there is exactly one solution.
To find the number of solutions for a system of equations, you can use the method of elimination or substitution. Both methods involve manipulating the equations to eliminate one of the variables, which will allow you to solve for the remaining variable and determine the number of solutions.
Let's go through the process step by step for the given system of equations:
3x + 5y = 20 (Equation 1)
-2x + 5y = 20 (Equation 2)
Method 1: Elimination
To eliminate a variable, you need to multiply one or both equations by appropriate numbers such that when you add or subtract them, one of the variables cancels out. In this case, if you multiply Equation 2 by 3 and Equation 1 by 2, the x terms will cancel out.
6x + 10y = 40 (Equation 3)
-6x + 15y = 60 (Equation 4)
Adding Equation 3 and Equation 4 eliminates the x term:
(6x - 6x) + (10y + 15y) = (40 + 60)
25y = 100
Now, solve for y:
y = 100/25
y = 4
Substitute the value of y back into one of the original equations (e.g., Equation 1, since it has simpler coefficients):
3x + 5(4) = 20
3x + 20 = 20
3x = 0
x = 0/3
x = 0
Therefore, the solution to the system of equations is x = 0 and y = 4. Since there are unique values for both variables, the system has a single solution (a unique solution).
Method 2: Substitution
In this method, we isolate one variable in one equation and substitute its expression into the other equation.
From Equation 1:
3x + 5y = 20
Solve for x:
3x = 20 - 5y
x = (20 - 5y)/3
Substitute x in Equation 2 with this expression:
-2((20 - 5y)/3) + 5y = 20
Simplify Equation 2:
-40/3 + (10y/3) + 5y = 20
Combine like terms:
-40 + 10y + 15y = 60
25y = 100
Solve for y:
y = 100/25
y = 4
Now, substitute the value of y back into x = (20 - 5y)/3:
x = (20 - 5(4))/3
x = (20 - 20)/3
x = 0/3
x = 0
Again, the solution to the system of equations is x = 0 and y = 4. As before, there is a single solution.
By using either the elimination or substitution method, you can quickly determine the number of solutions for a system of equations. If the coefficients of both variables are the same, it signifies an infinite number of solutions (a dependent system). If the coefficients are different, it indicates a single solution (an independent system).