Solve the equations 1, 3 / x + 1 = 3 / 5x-7 2, 7Y = 5x + 11

Solve the equations 1, 3 / x + 1 = 3 / 5x-7 2, 7Y = 5x + 11

Multiply both sides by 3 to get x + 1 = 5x-7
The result is - 4x = - 8
X=2
Substituting x = 2 into formula 2, we get 7Y = 10 + 11
Y=3

If 7Y = 1 / 5x, x =, y=

X=35y Y=x/35

Adding or multiplying two real roots of quadratic equation with one variable
If the two roots of the quadratic equation AX ^ 2 + BX + C = 0 (a is not equal to 0) are x1, X2, then X1 + x2 =?, x1 · x2 =?

If ax ^ 2 + BX + C = 0 (a is not equal to 0) has two,
Then the product of the two is C / A and the sum is - B / A
Namely: X1 + x2 = - B / A
x1·x2=c/a

The root formula of quadratic equation with one variable

Mathematical quadratic equation of one variable,
In isosceles △ ABC, BC = 8, the lengths of AB and AC are two of the equations x ^ 2-10x + M = 0 about X, then the value of M is
I'm sorry to do it in the way of quadratic equation of two variables. It's not AB, the length of AC is the equation of X, it's AB, the length of BC is the equation of X, wrong number, sorry

① If AB = AC, then △ = 0 (- 10) &# 178; - 4 × 1 × M = 0100-4m = 0m = 25, substituting M = 25 into the original equation, we get X & # 178; - 10x + 25 = 0 (X-5) &# 178; = 0, x = 5, AB = AC = 55 + 5 ＞ 8. ② if AB = BC = 8, then the original equation is 64-80 + M = 0m = 16, substituting M = 16 into the original equation, we get X & # 178; - 10x + 16 = 0x1 = 2 X

Mathematical problems of quadratic equation with one variable,
In the right angle △ ABC, ∠ C = 90 °, AC = 3, BC = 4, point E is on AC, point F is on AB (neither point e nor f coincides with the end point), and the line EF bisects the perimeter and area of the right angle △ ABC at the same time to find the length of AC?
For AE,

Using analytic geometry
Let e (0, b) be equal in area, f (6 / (3-B), 3 - (9 / 6-2b))
According to the equal perimeter, AE + AF = 6, B = (radical 6) / 2
So AE = 3 - (radical 6) / 2

Write a quadratic equation of one variable whose sum of two real roots is 2

x²-2x=0

Ask questions to find the unknown X: 0.4 = 13.3:1 and 20.75: x = 2.25:2:1, x = 1:30%
x: 0.4 = 13.3:1 and 20.75: x = 2.25:1,4: x = 1:30%

x: 4 = 13.3:1 and two fifths
x=13.3x0.4÷1.4
x=3.8
75: x = 2.25:1 / 2
x=0.75x0.5÷2.25
x=1/6
4 / x = 1 / 3:30%
x=1/3x4÷3/10
x=40/9
If you don't understand this question, you can ask,

"- 2 ≤ a ≤ 2" is the ()
A. Necessary and insufficient condition B. sufficient and unnecessary condition C. sufficient and necessary condition D. neither sufficient nor necessary condition

If the quadratic equation x2 + ax + 1 = 0 with real coefficients has no real roots, then △ = A2-4 ＜ 0, ∧ - 2 ＜ a ＜ 2, ∧ - 2 ≤ a ≤ 2, ∧ "- 2 ≤ a ≤ 2" is a necessary and insufficient condition that the quadratic equation x2 + ax + 1 = 0 with real coefficients has no real roots

Are 4 / A + B and 7 / x + y monomials

No, monomials can only have one unknown