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.
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
4 answers
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.
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.
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
5x+y=9
10x-7y+-18
1
-1
2
-2
7
-7