1.) Write a system of equations that has NO solution.
2.) If you wanted to eliminate the y variables in the system of equations below, what would you multiply the first and second equations by?( Do not solve the equation) 2x-3y=8
7x-4y=11
Try
x+t=14
2x+2y=28
2. If you multiply #1 by 4 the coefficient of y will be -12. Multiply #2 by 3 will give the coefficient of y as -12. \Subtracting then would eliminte y.
Check my thinking.
2.)
Multiply the first equation by 4 and the second equation by -3.
4(2x - 3y = 8), so 8x - 12y = 32
-3(7x - 4y = 11), so -21x + 12y = -33
So, the first equation has -12y and the second equation has 12y, so the y variables cancel out.
Try
x+t=14
2x+2y=28
I made a typo. first equation should be x+y=14 y and not t.
Note that Adams' response eliminates y by addition in problem #2. My response eliminates y by subtraction. Either is correct.
What if you wanted to eliminate the 'y'? What would you multiply the first equation by?
5x+y=9
10x-7y+-18
1
-1
2
-2
7
-7
1.) To create a system of equations with no solution, we need to ensure that the two equations represent parallel lines. This can be achieved by multiplying one equation by a constant that is not equal to the ratio of the coefficients of the other equation.
Let's consider the following system of equations:
Equation 1: 3x - 2y = 5
Equation 2: 6x - 4y = 10
If we try to solve this system by the method of substitution or elimination, we will find that the two equations are equivalent. In other words, they represent the same line. This means that they have an infinite number of solutions, but not no solution.
To modify the system so that it has no solution, you can introduce a contradiction by multiplying one of the equations by a value that is equal to the ratio of the coefficients of the other equation.
Let's rewrite the first equation as:
Equation 1: 3x - 2y = 5
Now, we will multiply the second equation by the ratio of the coefficients of the first equation, which is 6/3 (or 2):
Equation 2: (6/3)(6x - 4y) = (6/3)(10)
Simplifying Equation 2, we get:
Equation 2: 12x - 8y = 20
Now, if we try to solve the modified system using the method of substitution or elimination, we will find that the equations are contradictory, and there is no solution.
2.) To eliminate the y variables in the system of equations 2x-3y=8 and 7x-4y=11, we can multiply both equations by a suitable constant. The aim is to make the coefficients of the y terms the same or a multiple of each other. This allows us to add or subtract the equations to eliminate the y variables.
In this case, to eliminate the y variables, we need the coefficients of the y terms to be -3 and -4 (or multiples of each other). The least common multiple of -3 and -4 is 12. Thus, we need to multiply the first equation by 4 and the second equation by 3, or vice versa.
By multiplying the first equation by 4 and the second equation by 3, we get:
Equation 1: 4(2x - 3y) = 4(8)
Equation 2: 3(7x - 4y) = 3(11)
Simplifying these equations, we get:
Equation 1: 8x - 12y = 32
Equation 2: 21x - 12y = 33
Now, notice that the y variables have coefficients of -12 in both equations. This allows us to add or subtract the equations to eliminate the y variables, depending on the form required for solving the system further.