Question

Решите систему.a){12x-7y=2; {4x-5y=6.б){7u+2v=1; {17u+6v=-9. в){6x=25y+1; {5x-16y=-4. Реклама if (!window.hideDisplayAds) { googletag.cmd.push(function() { googletag.display('gpt-ad-brainly_rectangle_in_question'); }); } Попроси больше объяснений СледитьОтметить нарушение Tach4242 06.05.2012

Answer 1

a)(4x-5y=6

   (12x-7y=2

   

(икс=шесть четвертых - минус пять четвертых игрек

(12x-7y=2

 

(икс=шесть четвертых - минус пять четвертых игрек

(18-12y=2

 

(икс=шесть четвертых - минус пять четвертых игрек

(12y=16

 

(y=четыре третьих

(x= минус одна третья

Answer 2

Solution to System of Equations

Let's solve the given system of equations:

a) {12x-7y=2 {4x-5y=6

b) {7u+2v=1 {17u+6v=-9

c) {6x=25y+1 {5x-16y=-4

To solve these systems of equations, we can use various methods such as substitution, elimination, or matrix methods. Let's solve each system using the substitution method.

Solving System a)

We have the following equations: {12x-7y=2 (Equation 1) {4x-5y=6 (Equation 2)

From Equation 1, we can express x in terms of y: 12x = 2 + 7y x = (2 + 7y)/12

Substituting this value of x into Equation 2, we get: 4((2 + 7y)/12) - 5y = 6 (2 + 7y)/3 - 5y = 6 2 + 7y - 15y = 18 -8y = 16 y = -2

Substituting the value of y back into Equation 1, we get: 12x - 7(-2) = 2 12x + 14 = 2 12x = -12 x = -1

Therefore, the solution to System a) is x = -1 and y = -2.

Solving System b)

We have the following equations: {7u+2v=1 (Equation 3) {17u+6v=-9 (Equation 4)

From Equation 3, we can express u in terms of v: 7u = 1 - 2v u = (1 - 2v)/7

Substituting this value of u into Equation 4, we get: 17((1 - 2v)/7) + 6v = -9 (17 - 34v)/7 + 6v = -9 17 - 34v + 42v = -63 8v = -80 v = -10

Substituting the value of v back into Equation 3, we get: 7u + 2(-10) = 1 7u - 20 = 1 7u = 21 u = 3

Therefore, the solution to System b) is u = 3 and v = -10.

Solving System c)

We have the following equations: {6x=25y+1 (Equation 5) {5x-16y=-4 (Equation 6)

From Equation 5, we can express x in terms of y: 6x = 25y + 1 x = (25y + 1)/6

Substituting this value of x into Equation 6, we get: 5((25y + 1)/6) - 16y = -4 (125y + 5)/6 - 16y = -4 125y + 5 - 96y = -24 29y = -29 y = -1

Substituting the value of y back into Equation 5, we get: 6x = 25(-1) + 1 6x = -25 + 1 6x = -24 x = -4

Therefore, the solution to System c) is x = -4 and y = -1.

In summary, the solutions to the given systems of equations are: a) x = -1, y = -2 b) u = 3, v = -10 c) x = -4, y = -1

Please let me know if you need any further assistance!