solve by elimination method:2r +3s=29 and 3r +2s=16

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To solve the system of equations using the elimination method, we will first multiply the equations by a constant so that the coefficients of one of the variables will cancel out when we add the equations together.

Let's start by multiplying the first equation by 3 and the second equation by 2:

(3 * (2r + 3s) = 3 * 29) -> 6r + 9s = 87 Equation (1)
(2 * (3r + 2s) = 2 * 16) -> 6r + 4s = 32 Equation (2)

Now we have the following system of equations:

Equation (1): 6r + 9s = 87
Equation (2): 6r + 4s = 32

Next, we will subtract Equation (2) from Equation (1) to eliminate the 'r' variable:

(6r + 9s) - (6r + 4s) = 87 - 32
6r - 6r + 9s - 4s = 87 - 32
0r + 5s = 55
5s = 55

To solve for 's', divide both sides of the equation by 5:

5s/5 = 55/5
s = 11

Now that we have the value of 's', substitute it back into one of the original equations, let's use Equation (1):

6r + 9s = 87
6r + 9(11) = 87
6r + 99 = 87
6r = 87 - 99
6r = -12

To solve for 'r', divide both sides of the equation by 6:

6r/6 = -12/6
r = -2

Therefore, the solution to the system of equations is r = -2 and s = 11.

Multiply eqn 1 by 2 and eqn 2 by -3, then add the two equations to eliminate s. Then solve.

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