• R=1y There are three Rs in 15Z and in 3Y so there is one R in 1Y. Or R=Y
  • 15z = 3y Divide each side of the equation by three. 5z = y Looks familiar right? Since you already know that r = 5z, you can substitute r for 5z. Therefore: r = y
  • There are several ways to solve these. If you have equations like this: x + 2y = 5 2x + 3y = 8 Method 1: solve one equation for one unknown, in terms of the others, and substitute into the other equations: say I solve for x in the second equation: 2x + 3y = 8 so x = 4 - (3/2)y Now substitute this value for x into the other equation(s): 4 - (3/2)y + 2y = 5 You have one fewer equation to solve in one less unknown. Eventually you end up with one equation in one unknown like this, which is easily solved and you work backwards to find the others. Method 2: Another way is by doing row operations. x + 2y = 5 2x + 3y = 8 If I multiply the whole of the first row by 3 and the whole of the second row by 2, the coefficients of y become the same: 3x + 6y = 15 4x + 6y = 16 Now I will subtract the whole of the first rwo from the second: 3x + 6y = 15 x + 0y = 1 Now I'll subtract three of the second row from the first: 0x + 6y = 12 x + 0y = 1 eventually I get equations like x + 0(of anything) = answer Method 3: Matrices as a matrix: (1 2)(x)=(5) (2 3)(y)=(8) The 2x2 matrix is multipled by the 2x1 matrix to give a 1x2 matrix. A X = B If I find the inverse to the matrix A I can write: X = (inverse of A) B Then there are techniques for finding the inverse. (the row operations method is one)

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