solve for all values of x an y:system of equations 5x+2y=16 3x-5y=-9?

Question

solve for all values of x an y:system of equations 5x+2y=16 3x-5y=-9?

Answer 1

OK, so for the solution to a system, we are looking for the spot where the two lines cross. I dont know what level of algebra you are taking, so I will attack this in the most basic way I can think of.

first, line the systems up so they are pretty

5x+2y=16

3x-5y=-9

we could solve for y on both equations, and set those ugly things equal..but there would be a mess of nasty fractions that way. So..let's decide which variable to get rid of. The y terms have opposite signs. . so, I think getting rid of y is easiest. So, what is the smallest number that 2 and 5 go into? Yep! 10. so, to make the top line have 10y, we want to multiply through by 5. and for the bottom line, we want to go through by 2

5(5x+2y)=(5)16===>25x+10y=80

2(3x-5y)=2(-9)====> 6x-10y=-18

NOW..add the two lines together.. the y's knock each other out, and we just need to find x.

31x=62

now, divide both sides by 31 to get x by itself->

x=2



OK, we found x!! Now, we can either repeat the above process to clear x's..or just stick a 2 in where we see x...(I think this is easier)



so..1st equation.. since x=2, we get

5(2)+2y=16

10+2y=16 (Now, subtract 10 from both sides)

2y=6 (and divide by 2)

y=3

Voila!! We found X and Y. BUT..it is always a good idea to double check..so now, plug both values into the second equation to make sure it works

x=2, y=3

3x-5y=-9 becomes 3(2)-5(3)=-9==>6-15=-9==>-9=-9 (since both sides of the equal sign match, that means we found the right answer)



This method can help you with any systems...(sometimes the numbers get big and ugly, but it will always work)



Now, try a few on your own and see how it comes out!

Answer 2

a) 5x + 2y = 16. Find two numbers that add up to 16. One divisible by 5 and the other by 2 (10 and 6). Then divide 10 by 5 to give you x (2) and divide 6 by 2 to give y (3). x = 2, y = 3.

b) 3x - 5 y = -9. Find two numbers where subtracting one from the other will give you -9. The first must be divisble by 3 and the second by 5. (6, 15) Then divide 6 by 3 to give you x and 15 by 5 to give you y. x = 2, y =3

Answer 3

5x + 2y = 16

3x - 5y = -9



multiply the first equation by 5

25x + 10y = 80

multiply the second equation by 2

6x - 10y = -18



25x + 10y = 80

6x - 10y = -18

-------------------

31x = 62

x = 2



5x + 2y = 16

5(2) + 2y = 16

10 + 2y = 16

2y = 6

y = 3

(2, 3)



checking

5x + 2y = 16

3x - 5y = -9



5(2) + 2(3) = 16

10 + 6 = 16

16 = 16



3(2) - 5(3) = -9

6 - 15 = -9

-9 = -9

Answer 4

5x + 2y = 16

5x = 16 - 2y

x = (16 - 2y)/5

3x = 3(16 - 2y)/5



3x - 5y = -9

3(16 - 2y)/5 - 5y = -9

5[(48 - 6y)/5 - 5y] = 5(-9)

48 - 6y - 25y = -45

-31y = -45 - 48

-31y = -93

y = -93/-31

y = 3



5x + 2y = 16

5x + 2(3) = 16

5x = 16 - 6

x = 10/5

x = 2



∴ x = 2 , y = 3

Answer 5

5X+2Y=16) +(3X-5Y)=-9. = 8X-3Y=7(eq 3)

in eq (1) Y=(16-5X)/2. subs in Eq 3.

8X-3(16-5X)/2 =7: 8X-(48+15X)/2=7

(16X-48+15X)/2=7

16X-48+15X=14; 16X+15X=14+48=62; 31X=62

X=62/31=2. subs value of X in eq(1)5X+2Y=16

10+2Y=16 or 2Y=16-10=6

Y=6/2=3

Answer 6

5x +2y= 16 (1)

3x- 5y= -9 (2)



multiply eqn.(1)*3 & eqn.(2)*5

15x +6y= 48 (3)

15x- 25y= -45 (4)



subtract 15x from both eqn.s

6y= 48---> y= 8

-25y= -45---> y= 1.8



substitute y values into eqn.(1)

5x +2y= 16

5x +2(8)= 16 ---> x= 0

5x +2(1.8)= 16---> x= 2.48



(0,8) & (2.48, 1.8)

Answer 7

5X = 16-2Y ==> X = (16-2y) / 5



3 (16-2Y) / 5 - 5Y = -9

48-6Y / 5 - 5Y = -9

48-6Y - 25Y = -45

93 = 31Y

Y = 3



IF Y=3, THEN X = 2

Answer 8

25x + 10y = 80

6x - 10y = - 18--------ADD



31x = 62

x = 2



10 + 2y = 16

y = 3



x = 2 , y = 3

Answer 9

Since I assume this is for homework, I will tell you how to do a similar problem using the method I would use, instead of giving you the answer.



Let's use the system



1) 6x+5y=8

2) 4x-2y=2



Since each is an equation, you can manipulate a single equation on its own and still maintain equality. The key to a system such as this is to eliminate all variables but one. The way to do that is to find the lowest common multiple of the coefficient of the variable you are trying to eliminate. In my system let's eliminate x. Find the lowest common multiple of 4 and 6. It's 12. So you multiply both sides of equation 1 by 2 leaving you with the modified yet still equal equation 12x+10y=16

Multiply equation 2 by 3 so that your modified equation is 12x-6y=6



Then it's basically a subtraction problem of equations.



(12x+10y=16)

-(12x-6y=6)

-----------------------

0x +16y = 10



Thus, y= 10/16=5/8



Then to solve for y, just plug the number that was just solved for back into either original equation.



4x-2(5/8)=2

4x-5/4=2

4x=13/4

x=13/16



As a sanity check, plug both numbers into the other unused equation to see if the equality still holds.



6*(13/16) +5*(5/8)=8

39/8+25/8=8

64/8=8

8=8 (see, it worked!)





That's the basic method for solving these systems of equations.