Solve this system of linear equation. separate the x- and y-values with comma. 16x=6-18y -10x=-24+18y

(x,y) = (-3,3)

Step-by-step explanation:

16x=6-18y

-10x=-24+18y

We need to solve the above equations to find the values of x and y.

Rearranging the above equations

16x + 18y - 6 = 0 => eq(1)

-10x -18y +24 =0 => eq(2)

Adding eq(1) and eq(2)

16x + 18y - 6 = 0

-10x -18y +24 =0

6x + 18 =0

6x = -18

x= -18/6

x= -3

Putting value of x in eq(1) i.e

16x + 18y - 6 = 0

x= -3

16(-3) + 18y - 6 =0

-48 + 18y -6 =0

18y - 54 = 0

18y = 54

y= 54/18

y= 3

So, value of x = -3 and y = 3

(x,y) = (-3,3)

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3