Solving the equations 2S + 3T = - 1 4s-3t = 8

Solving the equations 2S + 3T = - 1 4s-3t = 8

2s+3t=-1（1）
4s-3t=8（2）
（1）+（2）
6s=7
s=7/6
（1）×2-（2）
6t+3t=-10
9t=-10
t=-10/9
therefore
s=7/6
t=-10/9

The solution of 4S + 3T = 5 2s-t = - 5

4S+3t=5.（1
2s-t=-5.（2
2) 2
4S－2t＝－10.(3
1) - 3) the results are as follows:
5t＝15
So t = 5
Substituting t = 5 into 2), we get the following results:
2S－5＝－5
So s = 0
So the solution of the original equations is:
｛S＝0
｛t＝5
For reference! Jswyc

Equations 2S + 3T = 1, 4s-9t = 8

2s+3t=1.(1)
4s-9t=8.(2)
(1)*2-(2) 6t+9t=2-8,15t=-6
t=-2/5
Substituting (1) 2S + 3 * (- 2 / 5) = 1,2s = 11 / 5
s=11/10

Factorization: (a + 1) ^ (2a-3) - 2 (a + 1) (3-2a) + 2a-3

：（a+1）^2*（2a-3）-2（a+1）（3-2a）+2a-3
=（2a-3)[(a+1)^2+2(a+1）+1）】
=（2a-3)*(a+1+1)^2
=(2a-3)(a+2)^2

Factorization of B (a + b) ^ 2 + 2B (a + b) - 2Ab ^ 2

Factorization of (a + b) (a + 2Ab + b) + A ^ 2B ^ 2-1

Original expansion = A2 + 2a2b + AB + AB + 2ab2 + B2 + a2b2-1
=(a+b)2+2ab(a+b)+(ab)2-1
=[(a+b)+ab]2-1
=(ab+a+b+1)(ab+a+b-1)

Factoring a ^ 4-A ^ 3-A ^ 2 process, thank you, quick

a^4-a^3-a^2
=a²(a²-a-1)

It is known that the value of the algebraic formula (x of Y-X of Y) divided by (x + y) is equal to 0

Solution;
(x/y-y/x)=(x²-y²)/xy
(y of x-x of Y) divided by (x + y)
=(x-y)/xy=0
∴
x=y

The sum of three times of X and two times of Y divided by the difference between two times of X and three times of Y is expressed as

(3x+2y)/(2x-3y)

It is known that the quadratic function y = ax & # 178; + 2x + 1 satisfies y ＞ 0, and the value range of a is obtained

The solution is established by the quadratic function y = ax ^ 2 + 2x + 1 satisfying Y > 0
Know a > 0 and Δ < 0
That is, a > 0 and 2 ^ 2-4a < 0
That is, a > 0 and a > 1
The solution is a > 1