Several mathematical problems of linear equation of three variables 1. There are three kinds of goods: A, B and C. If you buy 4 pieces of a, 5 pieces of B and 1 piece of C, it costs 230 yuan in total; if you buy 7 pieces of a, 9 pieces of B and 1 piece of C, it costs 385 yuan in total? 2. Given that X and y satisfy 2x-y = 3M, x + 2Y = 4m + 5, and X + y = 0, find M

Several mathematical problems of linear equation of three variables 1. There are three kinds of goods: A, B and C. If you buy 4 pieces of a, 5 pieces of B and 1 piece of C, it costs 230 yuan in total; if you buy 7 pieces of a, 9 pieces of B and 1 piece of C, it costs 385 yuan in total? 2. Given that X and y satisfy 2x-y = 3M, x + 2Y = 4m + 5, and X + y = 0, find M

It costs 230 yuan to buy 4 pieces of a, 5 pieces of B and 1 piece of C
It costs 460 yuan to buy 8 pieces of a, 10 pieces of B and 2 pieces of C
It costs 385 yuan to buy 7 pieces of a, 9 pieces of B and 1 piece of C
One piece for each of a, B and C = 230 × 2-385 = 75
2x-y = 3M, x + y = 0 -------- subtraction of two formulas, x = m, y = - M
Substituting x + 2Y = 4m + 5,
m-2m=4m+5
m=-1

A simple mathematical problem with additional points
2x＋3y＋z＝1
2x+y+z=-2
3x-2y-z=-4
Give points according to time
come on.

Let's set those three equations to 1.2.3
From formula 1-2, 2Y = 3, then y = 3 / 2
Then from formula 2 + 3, we can get 5x-y = - 6
Substituting a known y value
Get X
Finally, the solution is obtained

How to solve a number of mathematical problems
1.【3x-5y+2z=1 4x+5-z=5 x-5y+3z=-2
2.【6x+7y+4z=20 3x+4y+2z=10 2x+6y+7z=18
3.【2x+3y-4z+-2 3x-2y+5z=26 2x+5z=31

The three equations in each problem are sorted respectively, marked as ①, ②, ③ 1, 4x-z = 0, z = 4x from ②, 11x-5y = 1, 13x-5y = - 2, x = - 3 / 2, z = - 6, y = - 7 / 22, ② × 2 - ①, y = 0, 6x + 4Z = 20, 2x + 7z = 18 from ①, ③, and x = 2, z = 2.3 from ③

There is a formula that 7 / 8 times a = 6 / 6 times b = 3 / 2 times C = 10 / 11 divided by D = 1. Please arrange the four numbers of ABCD from small to large ()
（ ）（ ）.

A = 8 / 7 = 1 and 1 / 7
B=6/6=1
C=2/3
D = 11 / 10 = 1 and 1 / 10 (1 / 10)
So c

Polynomial operation of one variable
1、 Problem description
Design a simple one variable sparse polynomial adder
2、 Basic requirements
The basic functions of the simple calculator include:
1. Input and build polynomials A and B in exponential ascending order
2. Calculate the sum of polynomial A and B, that is to establish polynomial a + B
3. Output polynomials a, B, a + B in exponential ascending order
3、 Tips and analysis
1. Polynomial of degree n: P (x, n) = P0 + p1x1 + P2X2 + +Every subitem of pnxn is composed of "coefficient" and "index", so it can be abstracted into a linear table composed of "coefficient, index pair", in which each item of polynomial corresponds to a data element in the linear table, In this case, there is no need to pay storage space to store it; based on this, we can use a single linked table with head node to represent a unary polynomial
For example, polynomials a = 3 + 6x3-2x8 + 12x20 and B = 2x-2-6x3 + 8x10 can be expressed as follows:
2. The definition of data type can be described as follows:
typedef struct pnode
{int coef; / * coefficient field*/
Int exp; / * exponential field*/
Struct pnode * next; / * pointer field, pointing to the next subitem whose coefficient is not 0*/
}PolyNode,*PolyLink;
Multilink a, B, C; / * polynomials a, B, C stored in single linked list*/
3. Basic function analysis
(1) Input polynomials to build a linked list of polynomials
First, create a single linked list with the leading node; then input the subitems with coefficients not 0 according to the order of exponential increase and a certain input format: "coefficient, index pair". Each subitem will establish a node, and insert it into the tail of the polynomial linked list. Repeat this until it stops when it meets the end of input flag, and finally generate an ordered linked list with exponential increase
(2) Polynomial addition
Polynomial addition rule: for two polynomials with the same exponent, the coefficients are added. If the sum of the coefficients is not zero, it will form one of the sum polynomials. For the terms with different exponents, it will directly form one of the sum polynomials
Add the two polynomials A and B represented by the single linked list in (1), and the result of the operation is to generate a new linked list using the original table space, and the representation and polynomial C. the operation rules are as follows:
Let the pointers PA and Pb point to the first node of the polynomial list a and B respectively
① If PA > exp < Pb > exp, the node indicated by PA is inserted into the "sum polynomial" list;
② If PA > exp > Pb > exp, the node indicated by Pb is inserted into the "sum polynomial" list;
③ If pa - > exp = = Pb - > exp, then calculate the coefficient and pa - > coef + Pb - > coef; if the sum is not zero, insert it into the "sum polynomial" list to delete the node indicated by Pb; otherwise, delete the node indicated by PA and Pb
Continue to compare the next item and repeat the above process until the end of a list in a and B. at this time, move the remaining nodes in the non empty list into and out of the "sum polynomial" list
(3) The output of polynomials
In the text interface, you can output polynomials in a way similar to mathematical expressions. For example, polynomial a can be displayed as:
A=3+6XÙ3-2XÙ8+12XÙ20
Note:
For example, the output form of sub item 1x8 is X8, and the output form of item - 1x3 is - X3
When the coefficient sign of the first term of the polynomial is positive, it does not output "+", other terms need to output "+", "-" signs

#include
#include
#define A1(a,b,c,d,e) gotoxy(a,b);printf("c%d=",d);scanf("%f",e);
#define A2(a,b,c,d,e) gotoxy(a,b);printf("e%d=",d);scanf("%f",e);
#Define B1 do {printf ("- do you want to continue (find function value / derivative) (Y / N)?")
#define B2 ch2=bioskey(0); printf("[%c]",ch2)
#define B3 if((ch2=='Y')||(ch2=='y')) tc(head1->next,head2->next,r); }
#define B4 while((ch2=='y')||(ch2=='Y'))
struct ma
{ float c;
float e;
struct ma *next; }
main()
{ struct ma *head1,*head2,*p,*q,*r,*rm;
float s1[100][3],s2[100][3],mid1,mid2,mid3;
int i,j,t,num1,num2;
char ch,ch1,ch2,ch11,ch21;
struct ma *add();
struct ma *mul();
void tc();
void output();
loop1: t=0; ch='y';
clrscr();
Printf ("please input the number of items of the first polynomial):";
scanf("%d",&num1);
q=head1=(struct ma*)malloc(sizeof(struct ma));
(*q).next=NULL;
if(num1>0)
{printf ("please input the coefficient (c) and exponent (E) of the first polynomial):");
for(i=1;ic); (*r).e=q->e; q=(*q).next;}
else if(p->ee) { (*r).c=p->c ; (*r).e=p->e; p=(*p).next; }
else { if(ch1=='+') (*r).c=(p->c)+(q->c);
else (*r).c=(p->c)-(q->c);
(*r).e=(p->e); p=(*p).next;q=(*q).next;}
}while((p!=NULL)&&(q!=NULL));
while(q!=NULL)
{ r=(struct ma*)malloc(sizeof(struct ma));
(*s).next=r; s=r;
if(ch1=='+') (*r).c=q->c ;
else (*r).c=-(q->c);(*r).e=q->e; q=(*q).next;}
while(p!=NULL)
{ r=(struct ma*)malloc(sizeof(struct ma));
(*s).next=r; s=r;
(*r).c=p->c ; (*r).e=p->e;p=(*p).next; }
(*r).next=NULL;
s=r=(*head3).next;
while(r->next!=NULL)
if(r->e==(r->next)->e)
{ r->c+=(r->next)->c; s=r->next;
r->next=(r->next)->next; free(s); s=r; }
else { r=r->next; s=s->next; }
s=r=(*head3).next;
return(r);
}
struct ma *mul(struct ma *head1,struct ma *head2)
{ struct ma *head3,*head4,*p,*q,*r,*s,*r1,*s1;
struct ma *w[100];
int t=0;
head3=r=s=(struct ma*)malloc(sizeof(struct ma));
head4=r1=s1=(struct ma*)malloc(sizeof(struct ma));
p=head1->next; q=head2->next;
while(p!=NULL)
{ s=(struct ma*)malloc(sizeof(struct ma));
s->c=(p->c)*(q->c);
s->e=(p->e)+(q->e);
r->next=s;
r=s;
p=p->next;}
s->next=NULL; s=r=head3->next; w[0]=head3;
p=head1->next; q=q->next;
while(q!=NULL)
{ while(p!=NULL)
{ s1=(struct ma*)malloc(sizeof(struct ma));
s1->c=(p->c)*(q->c);s1->e=(p->e)+(q->e);
r1->next=s1;
r1=s1;
p=p->next;
}
s1->next=NULL;r1=s1=head4->next;
t++;
w[t]=(struct ma*)malloc(sizeof(struct ma));
w[t]->next=add(w[t-1],head4,'+');
while(head3->next!=NULL) {r=r->next;free(s);head3->next=s=r;}
s=r=head3;
while(head4->next!=NULL) {r1=r1->next;free(s1);head4->next=s1=r1;}
s1=r1=head4;
p=head1->next; q=q->next;
}
return(w[t]->next);}
void tc(struct ma *p,struct ma *q, struct ma *r)
{ char ch1,ch2;
float x;
float qzhi(struct ma *head,float x);
void qdao();
Printf ("- please select the serial number of the polynomial:);
Printf ("\ n 1 -- first; 2 -- second; 3 -- third");
ch1=bioskey(0);printf("[%c]",ch1);
Printf ("- please input function code:);
Printf ("[n, Z -- function value, D -- derivative function");
ch2=bioskey(0);printf("[%c]",ch2);
switch(ch1)
{ case '1': { if(ch2=='z')
{printf ("- please enter the independent variable X:");
scanf("%f",&x);
Printf ("the value of polynomial is: 10.4f", qzhi (P, x));}
if(ch2=='d') qdao(p); break;}
case '2': { if(ch2=='z')
{printf ("- please enter the independent variable X:");
scanf("%f",&x);
Printf ("the value of polynomial is: 10.4f", qzhi (Q, x));}
if(ch2=='d') qdao(q); break;}
case '3': { if(ch2=='z')
{printf ("- please enter the independent variable X:");
scanf("%f",&x);
Printf ("the value of polynomial is: 10.4f", qzhi (R, x));}
if(ch2=='d') qdao(r); break;} } }
float qzhi(struct ma *head,float x)
{ float value=0;
while(head!=NULL)
{ value+=(head->c)*pow(x,head->e);head=head->next;}
return(value);
}
void qdao(struct ma *head)
{ struct ma *p,*q,*r,*t;
p=head;
t=r=q=(struct ma *)malloc(sizeof(struct ma));
while(p!=NULL)
{ q=(struct ma *)malloc(sizeof(struct ma));
r->next=q;r=q;
q->c=(p->c)*((p->e));q->e=(p->e)-1;
p=p->next;}
q->next=NULL;
Printf ("- the derivative of the polynomial is:);
output(t->next);
p=t;
while(t!=NULL) { p=t->next;free(t);t=p;}
}
void output(struct ma *r)
{ int l=0;
struct ma *t,*s,*head;
t=r;
while(r->next!=NULL)
if(r->e==(r->next)->e)
{ r->c+=(r->next)->c; s=r->next;
r->next=(r->next)->next; free(s); s=r; }
else { r=r->next; s=s->next; }
s=r=t;
head=(struct ma*)malloc(sizeof(struct ma));
head->next=r;
s=head;
while(r!=NULL)
{ if((r->c)==0) {s->next=r->next; free(r);r=s->next;}
else { s=s->next;r=r->next;}}
s=r=head->next;
while(r!=NULL)
{ if(r->c==1) { l++;
if(r->e==1) printf("x");
else if(r->ee);
else if(r->e==0) printf("1");
else printf("x^%.0f",r->e);}
else if(r->c==-1) { l++;
if(r->e==1) printf("-x");
else if(r->ee);
else if(r->e==0) printf("-1");
else printf("-x^%.0f",r->e);}
else { l++;
if(r->e==1) printf("%4.1fx",r->c);
else if(r->ec,r->e);
else if(r->e==0) printf("%4.1f",r->c);
else printf("%4.1fx^%.0f",r->c,r->e); }
if(((*r).next!=NULL)&&((r->next)->c>0)) printf(" +");
r=r->next; }
if(l==0) printf("0");
}

If the central angle of an arc is 72 degrees and the diameter of the central angle is 10 cm, the arc length is () cm
If it is equal to 1 / 4 of the circumference of the other circle, then the radius of the other circle is () cm

1、L=72/180×π×（10÷2）=6.28（cm）
2. Let R be the radius
2πr/4=2π
r=4
That is, the radius is 4cm

Collect information about education savings in this area and think about the following questions
(1) According to the way of education savings, 50 yuan can be saved every month for three consecutive years. How many yuan can the principal and interest be withdrawn at one time when it is due (three years) or six years?
(2) According to the way of education savings, we can deposit a yuan every month for three consecutive years. How many yuan can we withdraw the principal and interest at the time of maturity (three years) or six years?
(3) In the form of education savings, 50 yuan is deposited every month for three consecutive years. When it matures (three years), it can be withdrawn at one time with the same ratio of principal to interest
How much is the extra income of "zero deposit and lump sum withdrawal"?
(4) If you want to withdraw the principal and interest of education savings totaling 10000 yuan in three years, how many yuan should you deposit each month?
(5) I want to withdraw my education savings at one time after three years