solve by elimination method; 3r-7s=13, 7r-3s=69
To solve the system of equations using the elimination method, follow these steps:
Step 1: Multiply one or both equations by a constant. The goal is to make the coefficients of either the variable "r" or "s" the same in both equations. In this case, we can easily eliminate "s" by multiplying the first equation by 3 and the second equation by 7.
Multiplying the first equation by 3: 3 × (3r - 7s) = 3 × 13
Which gives us: 9r - 21s = 39 (Equation 1')
Multiplying the second equation by 7: 7 × (7r - 3s) = 7 × 69
Which gives us: 49r - 21s = 483 (Equation 2')
Step 2: Add the two equations (Equation 1' and Equation 2') together to eliminate the "s" variable.
(9r - 21s) + (49r - 21s) = 39 + 483
Combine like terms on the left side: 9r + 49r - 21s - 21s = 522
Simplify further: 58r - 42s = 522 (Equation 3)
Step 3: Solve Equation 3 for either "r" or "s". In this case, let's solve for "r".
To isolate "r", divide both sides of Equation 3 by 58:
(58r - 42s)/58 = 522/58
Simplify: r - (42s/58) = 9
Step 4: Solve for "s" using the value of "r" obtained from Step 3. Let's substitute the value of "r" from Equation 3 into either Equation 1' or 2'.
Let's substitute the value of "r" from Equation 3 into Equation 1':
9r - 21s = 39
Substitute: 9(9) - 21s = 39
Which gives us: 81 - 21s = 39
Step 5: Solve the equation obtained in Step 4 for "s".
Subtract 81 from both sides: -21s = 39 - 81
Simplify: -21s = -42
Divide both sides by -21 to solve for "s":
s = -42 / -21
s = 2
Step 6: Substitute the value of "s" (s = 2) into either Equation 1 or Equation 2 to solve for "r".
Let's substitute s = 2 into Equation 1:
3r - 7s = 13
Substitute: 3r - 7(2) = 13
Which gives us: 3r - 14 = 13
Add 14 to both sides: 3r = 13 + 14
Simplify: 3r = 27
Divide both sides by 3 to solve for "r":
r = 27 / 3
r = 9
So the solution to the system of equations is r = 9 and s = 2.