I want to ask a few questions, is it a system of equations, thank you, I hope to hurry up o oh urgen

Question 1 3x+4y=16

5x-6y=33

So 15x+20y=80... Type 1.

15x-18y=99。。。Type 2.

Subtract formula 1 from formula 2 to get 38y = minus 19

So y = minus 1 2 and x = 6

Question 2. Fractions are reduced to integers, resulting in 16x+12y=minus 5... Type 1.

3x-2y=minus 2... Type 2.

Multiply 2 by 6 and add 1 to get 34x = minus 17

So x = minus 2 and y = minus 2

Question 3. Half x-y = 6... Type 1.

2x+31y=minus 11... Type 2.

Multiply the formula by 4 and subtract the formula 2 to get minus 35y=35

So y = minus 1 and x = 10

One, 3x+4y=16 5x-6y=33

The first formula sub-shifts the term to get 4y=16-3x

Divide the two sides by 4 to get y=4-3x 4

Substituting y into the second equation gives 5x-6 (4-3x 4)=33 to get x=6 and substituting x=6 into 1 to get y=-1 2 two, 2x 3+y 2=-5 24

x/4-y/6=-1/6

Multiply both sides of Eq. 1 and 2 by 24 to give 16x+12y=-56x-4y=-4

Multiply 2 by 3 and then add 1 to get 34x=-17

Get x=-1 2

y=-1 4

Three, x 2-y = 6

2x+31y=-11

Multiply 1 by 4 to get 2x-4y=24 minus 2.

Get -35y=35

Get y=-1 into 1 to get x 2+1=6

Get x=10

1. 3x+4y=16 multiply each side of the equation by 5, 5x-6y=33 multiply each side by 3, and then get the two equations to subtract to get y=, substitute y= into 3x+4y=16, and find x=6

2. Remove the denominator of the two equations to get two equations: 8x+12y=--5, 6x-4y=-1, multiply the two sides of 6x-4y=-1 by 3 to get 18x-12y=-3, and then add it with the equation 8x+12y=--5 to obtain x=minus four-tenths, and substitute x=minus four-thirteenths into any equation to obtain y=minus fifty-twoths eleven.

3. Multiply the two sides of the equation one-half x-y=6 by 4 to get 2x-4y=24, and combine it with the equation 2x+31y=minus 11 to find y=-1, and then substitute y=-1 into any equation to obtain x=10

Using the substitution method, y is represented by x and the first equation is brought in.

If there is no problem with the question you give, this question is just for you to play.

a p*a*ap=a*a*a=-512 (p and 1 p are eliminated), so a=-8, and then substitute the first equation to solve it.

Augmentation matrix (a, b) = [1 0 2 1] [1 a 5 0] [4 0 a+3 b+8] The elementary row transformation is [1 0 2 1] [0 a 3 -1] [0 0 a-5 b+4] then r(a) 2 (1) Since the system of equations ax=b has 2 different solutions, then r(a)=r(a,b)=2<3 （2） a=5, b=-4.The deformation of the same solution of the system of equations is x1=1-2x3 5x2=-1-3x3 take x3=3, and the special solution (-5, -2, 3) t derives the group that the corresponding homogeneous equation is x1=-2x3 5x2=-3x3 takes x3=-5, and the basic solution system (10, 3, -5) t is obtained, then the general solution of the system of equations is x=(-5, -2, 3) t+k(10, 3, -5) t, where k is an arbitrary constant.

Multiply the first formula by 2 and subtract the second formula.

10x=680*2-1240=120 x=12

y=16

I can't see it clearly, but I can see it clearly.

x＋y=100 ①36x+20y=28*100②

by x=100-y

Substitute , then.

36*(100-y)+20y=28*100, y=50

Substituting y=50 into , we get x=50

x=50y=50