The following equations 9x + 8y = - 2,4x + 5Y = - 11 are solved by substitution method

The following equations 9x + 8y = - 2,4x + 5Y = - 11 are solved by substitution method

9x+8y=-2,(1)
4x+5y=-11 (2)
(2) 9 - (1) × 4
13y=-91
y=-7
x=(-11+5×7)/4
=6
Namely
x=6
y=-7

{9x = 8y = - 2 {4x = 5x = - 11 and solved by substitution method

Is it 9x + 8y = - 2 (1)
4x+5x=-11 (2)?
By (1)
8y=-2-9x
y=(-2-9x)/8
Substituting (2)
4x+5(-2-9x)/8=-11
Take eight on both sides
32x-10-45x=-88
13x=78
x=6
y=(-2-9x)/8=-7

Solve the equations {9x + 5Y = 134x-13y = - 2

9x + 5Y = 13 or so, with the remaining 4 as follows:
36X+20Y=52 （a）
4x-13y = - 2 and multiply by 9 to get:
36X-117Y=-18 (b)
(a) (b) 137y = 70
Y=70/137
Take y = 70 / 137 into (a) to get x = 159 / 137
Hand only,

Using the formula (a + b) (a-b) = A2-B2 to calculate (x + 2y-1) (x-2y + 1), the following deformation is correct ()
A. [x-（2y+1）]2B. [x+（2y+1）]2C. [x-（2y-1）][x+（2y-1）]D. [（x-2y）+1][（x-2y）-1]

(x + 2y-1) (x-2y + 1) = [x - (2y-1)] [x + (2y-1)]

The sum of the two numbers is 6 and the difference is 8. The quadratic equation of one variable with the two numbers as roots and the coefficient of quadratic term as 1 is

The sum of the two numbers is 6 and the difference is 8. The quadratic equation of one variable with the two numbers as roots and the coefficient of quadratic term as 1 is
x1+x2=6
x1-x2=8
The solution is X1 = 7, X2 = - 1
Because the coefficient of quadratic term is 1, i.e. a = 1, let the quadratic equation of one variable be
x ²+bx+c=0
Substituting X1 = 7, X2 = - 1
7² + 7b + c =0
（-1）²+（-1）x b + c = 0
The solution is b = - 6 and C = - 7
So the equation is X & # 178; - 6x-7 = 0

3 △ 1 / 2 x = 5 / 6 x!
3÷（1÷2）x=5÷6

3÷1/2X=5/6
6X=5/6
X=5/6 * 1/6
X=5/36

The quadratic equation of one variable with roots of 2, - 4 is (the coefficient of quadratic term is 1)

x²+bx+c=0
x=2,x2=-4
Then X1 + x2 = - 2 = - B
x1x2=-8=c
So x & # 178; + 2x-8 = 0

1.5x + 5 × 0.6 = 12 the solution of equation x is unknown

1.5x+5×0.6=12
1.5x+3=12
1.5x=12-3
1.5x=9
x=9/1.5
x=6

What is the univariate quadratic equation with roots - 3 and 7 and quadratic coefficient 1?

From the relationship between root and coefficient, we can get: Tan α + 1 / Tan α = k Tan α * 1 / Tan α = (3K ^ 2-13) / 3 and Tan α * 1 / Tan α = 1. The solution equation 1 = (3K ^ 2-13) / 3 is due to 3 π & lt; α & lt; 7 π / 2

2.5x-0.6 * 3 = 0.7 for unknowns

2.5x-1.8=0.7
2.5x=2.5
x=1