Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution.
5/4x − 2/3y = 8
1/4x + 5/3y = 7
Find the solution, if one exists.
There are technically 2 different ways that you could go about solving this problem. The first option is definitely the easier way to solve this problem because it is completely visual - unless you prefer to solve things algebraically. The first way to solve this problem would be just to graph the two equations in a graphing calculator. (If you do not have one, I would reccomend using the wesite Desmos. It is a free online graphing calculator.)
When you graph the two equations, look to see how many times they intersect or meet at the same point. If this only occurs once, then there is only one solution. If they meet at two points, then there are two solutions. If the 2 equations overlap each other when they are graphed, then there are infinitely many solutions. If the 2 equations do not touch at all on the graph, then there is no solution. In this case, when you graph the 2 eqations you only see one place where both equations meet at the same point. That point is (8,3).
To determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution, we need to solve the system of equations.
The system of equations is:
5/4x - 2/3y = 8 ...(Equation 1)
1/4x + 5/3y = 7 ...(Equation 2)
To eliminate the fractions, we can multiply both sides of Equation 1 by the least common multiple (LCM) of the denominators, which is 12, and multiply both sides of Equation 2 by the LCM as well.
Multiplying Equation 1 by 12, we get:
(12)(5/4x) - (12)(2/3y) = (12)(8)
15x - 8y = 96 ...(Equation 3)
Multiplying Equation 2 by 12, we get:
(12)(1/4x) + (12)(5/3y) = (12)(7)
3x + 20y = 84 ...(Equation 4)
Now, we have the following system of equations:
15x - 8y = 96 ...(Equation 3)
3x + 20y = 84 ...(Equation 4)
To solve this system, we can use the method of elimination. We will multiply Equation 4 by 5 and Equation 3 by 3 to make the coefficients of y in both equations equal and eliminate y.
Multiplying Equation 4 by 5, we get:
5(3x + 20y) = 5(84)
15x + 100y = 420 ...(Equation 5)
Multiplying Equation 3 by 3, we get:
3(15x - 8y) = 3(96)
45x - 24y = 288 ...(Equation 6)
Now, we can subtract Equation 6 from Equation 5 to eliminate x:
(15x + 100y) - (45x - 24y) = 420 - 288
-30x + 124y = 132 ...(Equation 7)
We now have two equations:
-30x + 124y = 132 ...(Equation 7)
15x - 8y = 96 ...(Equation 3)
To eliminate x, we can multiply Equation 7 by 2 and Equation 3 by 4:
2(-30x + 124y) = 2(132)
-60x + 248y = 264 ...(Equation 8)
4(15x - 8y) = 4(96)
60x - 32y = 384 ...(Equation 9)
Now, we can add Equation 8 and Equation 9 to eliminate x:
(-60x + 248y) + (60x - 32y) = 264 + 384
216y = 648
Dividing both sides of the equation by 216, we get:
y = 3
Now, we can substitute y = 3 into Equation 3 to find x:
15x - 8(3) = 96
15x - 24 = 96
15x = 120
x = 8
Thus, the solution to the system of linear equations is:
x = 8
y = 3
Therefore, the system of linear equations has one and only one solution, which is x = 8, y = 3.
To determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution, we can use the method of solving systems of equations, such as elimination or substitution.
Let's use the method of elimination to solve this system of equations:
Given system:
1) (5/4)x - (2/3)y = 8
2) (1/4)x + (5/3)y = 7
First, we need to eliminate one variable by multiplying the equations by appropriate numbers to make the coefficients of one of the variables in both equations equal. In this case, we can multiply equation 1 by 3 and equation 2 by 4 to eliminate the y variable:
3) (15/4)x - (2/3)y = 24 (Multiply equation 1 by 3)
4) (4/4)x + (20/3)y = 28 (Multiply equation 2 by 4)
Simplifying the equations, we have:
3) (15/4)x - (2/3)y = 24
4) x + (20/3)y = 28
Now, subtract equation 4 from equation 3:
3) (15/4)x - (2/3)y - (x + (20/3)y) = 24 - 28
Simplifying the equation further:
(15/4)x - (2/3)y - x - (20/3)y = -4
Combining like terms:
(15/4 - 1)x + (-(2/3) - (20/3))y = -4
Simplifying further:
(11/4)x - (22/3)y = -4
Now, we have a new equation:
5) (11/4)x - (22/3)y = -4
This equation is consistent with the original system of equations. To find the solution, we need to solve this equation. However, since we still have two variables, x and y, we need another equation to determine their specific values.
To determine the solution, we will need a third equation. Without a third equation, we cannot find the exact values for x and y. Therefore, this system of equations does not have a unique solution.
In conclusion, the system of linear equations:
1) (5/4)x - (2/3)y = 8
2) (1/4)x + (5/3)y = 7
Does not have one and only one solution. It either has infinitely many solutions or no solutions.