The solution of {mx-4y = 123x-2y = 8 is negative when m is a value I don't know what I did wrong. Here's my process mx-4y=12--(1) 3x-2y=8-----(2) (1) - (2) × 2 mx-6x=12-16 x=4/(6-m) Replace one of the equations and get y = [6 / (6-m)] - 4 Because x

The solution of {mx-4y = 123x-2y = 8 is negative when m is a value I don't know what I did wrong. Here's my process mx-4y=12--(1) 3x-2y=8-----(2) (1) - (2) × 2 mx-6x=12-16 x=4/(6-m) Replace one of the equations and get y = [6 / (6-m)] - 4 Because x

Replace one of the equations and get y = [6 / (6-m)] - 4 because X

Ten numbers from 0 to 9 are required to form an addition formula, and the numbers cannot be repeated

246+789=1035
249+786=1035
264+789=1053
269+784=1053
284+769=1053
286+749=1035
289+746=1035
289+764=1053
324+765=1089
325+764=1089
342+756=1098
346+752=1098
347+859=1206
349+857=1206
352+746=1098
356+742=1098
357+849=1206
359+847=1206
364+725=1089
365+724=1089
423+675=1098
425+673=1098
426+879=1305
429+876=1305
432+657=1089
437+589=1026
437+652=1089
439+587=1026
452+637=1089
457+632=1089
473+589=1062
473+625=1098
475+623=1098
476+829=1305
479+583=1062
479+826=1305
483+579=1062
487+539=1026
489+537=1026
489+573=1062
537+489=1026
539+487=1026
573+489=1062
579+483=1062
583+479=1062
587+439=1026
589+437=1026
589+473=1062
623+475=1098
624+879=1503
625+473=1098
629+874=1503
632+457=1089
637+452=1089
652+437=1089
657+432=1089
673+425=1098
674+829=1503
675+423=1098
679+824=1503
724+365=1089
725+364=1089
742+356=1098
743+859=1602
746+289=1035
746+352=1098
749+286=1035
749+853=1602
752+346=1098
753+849=1602
756+342=1098
759+843=1602
764+289=1053
764+325=1089
765+324=1089
769+284=1053
784+269=1053
786+249=1035
789+246=1035
789+264=1053
824+679=1503
826+479=1305
829+476=1305
829+674=1503
843+759=1602
847+359=1206
849+357=1206
849+753=1602
853+749=1602
857+349=1206
859+347=1206
859+743=1602
874+629=1503
876+429=1305
879+426=1305
879+624=1503

The chord length of line L with slope 2 cut by hyperbola x23 − Y22 = 1 is 4. The equation of line L is obtained

Let the equation of the straight line l be y = 2x + m and intersect the hyperbola at two points a and B. let the coordinates of two points a and B be a (x1, Y1) and B (X2, Y2) respectively. Substituting y = 2x + m into x23 − Y22 = 1, we can get: 10x2 + 12mx + 3 + 3 (M2 + 2) = 0, ｜ X1 + x2 = - 65m, x1x2 = 310 (M2 + 2) ｜ (x1-x2) 2 = (x1 + x2) 2-4x1x2 = 36m225-65 (M2 + 2) ｜ ab | 2 = (1 + K2) (x1-x2) 2 = 5 (x1-x2) 2 = 36m25-6 (M2 + 2) +2) The equation of the straight line is y = 2x ± 2103

Addition and subtraction of marks in mathematical problems

3/4+1/5=4/5 3/4-1/5=11/20 5/6+2/9=19/18 5/6-2/9=11/18 2/3+3/5=19/152/3-3/5=1/15 6/7+1/2=19/14 6/7-1/2=5/14 2/3+1/6=5/6 1/3+1/4=7/121/3-1/4...

For the quadratic equation AX2 + BX + C = 0, we prove that the two roots of the equation are X1 = 1, X2 = C / A
For the univariate quadratic equation AX2 + BX + C = 0, we know a + B + C = 0, and prove that the two roots of the equation are X1 = 1, X2 = C / A

If a + B + C = 0, then B = - a-c. if a + B + C = 0 is substituted into the original equation, then ax ^ 2 - (a + C) x + C = 0
ax^2-ax-cx+c=0 ax(x-1)-c(x-1)=0 (ax-c)(x-1)=0
x2=c/a,x1=1

a. B is a rational number, if | A-B | = a + B, then for the conclusion: ① a must not be negative; ② B may be negative, then the following judgment is correct ()
A. Only (1) correct B. only (2) correct C. All (1) correct D. all (1) incorrect

① When a ≥ B, A-B = a + B ｜ B = 0 ｜ a ≥ 0. ② when a ≤ B, B-A = a + B ｜ a = 0, B ≥ 0, so a and B are both nonnegative numbers, so a is selected

Solve the equation with Square Collocation: X & # 178; + 6x-7 = 0, and solve the equation with formula method: 2x & # 178; - 5x + 1 = 0

x²+6x-7=0
x²+6x+9-9-7=0
(x+3)²-16=0
(x+3)²=16
X + 3 = 4 or x + 3 = - 4
x1=1 x2=-7
2x²-5x+1=0
X1 = [5 + radical (- 5) ^ 2-4 * 2 * 1] / (2 * 2) = (5 + radical 17) / 4
X2 = [5-radical (- 5) ^ 2-4 * 2 * 1] / (2 * 2) = (5-radical 17) / 4

How to explain the multiplication and division of marks in sixth grade mathematics

Multiplication numerator and denominator are multiplied separately
Division is the multiplication of the numerator by the following denominator and the multiplication of the denominator by the following numerator

∫sint/(cost+sint)dt

∫sint/(cost+sint)dt= (1/2)∫[(sint+cost) +(sint-cost) ]/(cost+sint)dt=(1/2)∫ dt + (1/2)∫(sint-cost)/(cost+sint)dt=(1/2)∫ dt - (1/2)∫dln(sint+cost)=(1/2)t - (1/2)ln|sint+cost| + C

The sum of the two numbers is 30 and the difference is 12
Equation:__________________________
5 times of a number divided by 9 equals 15 to find a number
Equation:__________________________
The sum of the two brothers is 78 years old. My brother is 6 years older than my brother. How old is my brother?
Equation:__________________________

Let the decimal be x, then the large number is 30-x, 30-x-x = 12, so x = 9
Let a number be x, then 5x / 9 = 15, so x = 27
If the elder brother is x years old, then the younger brother is 78-x, X - (78-x) = 6, so x = 42