3r-5s=19 5r+3s=43 what is the solution of the System?

1. Make one of the variables cross out
In thi case multiply the top by 5 and bottom by -3
2. 15r-25s= 95
-15r-9s= -129
3. 15r cancel out so now add the other #'s

4. -34s= -34
divide by negative 34
{s=1}<-------answer for "s"

Now plug it in to solve for r(any eqaution)

3r-5(1)=19
3r-5= 19
add the 5
3r=24
divide by 3

{r=8}<-----answer for "r"

PLug it in to check

{8,1}<--------

Hope that helped

1st equation (s):
3r - 5s = 19
5s = 3r - 19
s = (3r - 19)/5

2nd equation (s):
5r + 3s = 43
3s = 43 - 5r
s = (43 - 5r)/3

Value of r:
3(3r - 19) = 5(43 - 5r)
9r - 57 = 215 - 25r
34r = 272
r = 8

Value of s (1st equation):
= (3[8] - 19)/5
= (24 - 19)/5
= 5/5 or 1

Answer: r = 8, s = 1

Proof (2nd equation):
5(8) + 3(1) = 43
40 + 3 = 43
43 = 43

(5r + 2t = 5) 7 --> to eliminate the variable `t` -(3r - 7t = 3) 2 --> to eliminate the variable `t` ____________ 35r + 14t = 35 -6r +14t = -6 _____________ 29r = 29 r = a million --replace the linked fee of `r` w/c is `a million` to any of the two unique equations-- 5(a million) + 2t = 5 5 + 2t = 5 2t = 5-5 t = 0 answer Set: {a million,0}

3r - 5s = 19 (solve by using elimination)
5r + 3s = 43

3r - 5s = 19
5(3r - 5s) = 5(19)
15r - 25s = 95

5r + 3s = 43
3(5r + 3s) = 3(43)
15r + 9s = 129

...15r - 25s = 95
‒) 15r + 9s = 129 (subtraction)
--------------------------------
-34s = -34

-34s = -34
s = -34/-34
s = 1

3r - 5s = 19
3r - 5(1) = 19
3r - 5 = 19
3r = 19 + 5
r = 24/3
r = 8

∴ r = 8 , s = 1

3r-5s=19
5r+3s=43

Using Cramer's rule
D = (3)(3) - (5)(-5) = 9 + 15 = 34
Dr = (19)(3) - (43)(-5) = 272
r = 272/34 = 8
Ds = 3(43) - 5(19) = 34
s = 34/34 = 1
(r, s) = (8, 1)

Let's add the wquations to produce a 3rd equation:

3r-5s=19
5r+3s=43

Let's multiple the 1st equation by 5 and the 2nd equation by 3 so we end up with r componants of the same quantity:
5( 3r - 5s = 19 )
15r - 25s = 95
Solve this for 15r
15r = +25s + 95
15r = 25s + 95

3( 5r + 3s = 43 )
15r + 9s = 129
Substitue in the value we calculated for 15r above
25s + 95 +9s = 129
25s + 9s = 129 - 95
34s = 34
s = 34/34 = 1

Substitute the value for s in the original 1st equation and solve for r
3r - 5s = 19
3r - 5(1) = 19
3r -5 = 19
3r = 19 + 5
3r = 24
r = 24/3 = 8

Sunstitute both values into the original 2nd equation to check:
5r + 3s = 43
5(8) + 3(1) = 43
40 + 3 = 43
43 = 43 That's true so our answers are correct

3r-5s=19 multiply by 5 --> 15r - 25s = 95
5r+3s=43 multiply by 3 --> 15r + 9s = 129

Subtract the second line (15r + 9s = 129) from the first line (15r - 25s = 95).
Then, -34s = -34
Therefore, s = 1
Apply s=1 into 3r-5s=19
3r - 5*1 = 19
3r - 5 = 19
3r = 24
r = 8

Answer is r = 8, s=1.

9r - 15s = 57
25r + 15s = 215-------ADD

34r = 272
r = 8

40 + 3s = 43
3s = 3
s = 1

r = 8 , s = 1