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This time you should notice that we don't have the same number of \(y\)'s.

We can get the same number of \(y\)'s if we multiply equation 1 by 4 and equation 2 by 3, so that both equations have \(12y\). (Also we need one 12y positive and one negative.)

\(8x + 12y = 64(Equation\,3)\)

\(9x - 12y = 21(Equation\,4)\)

\(8x + 12y = 64\)

\(9x - 12y = 21\)

\(17x = 85\)

\(x = \frac{{85}}{{17}} = 5\)

Substitute into equation 1: \(2x + 3y = 16\)

When \(x = 5\)

\(2 \times 5 + 3y = 16\)

\(10 + 3y = 16\)

\(3y = 16 - 10\)

\(3y = 6\)

\(y = \frac{6}{3} = 2\)

Therefore \(x = 5\) and \(y = 2\)

\(-4d - 2e = -4(Equation\,3)\)

\(15d+2c=4\)

\(-4d-2c=-4\)

\(11d = 0\)

\(d = 0 \div 11\)

\(d = 0\)

Substitute into equation 2: \(2d + e = 2\)

\(2 \times 0 + e = 2\)

\(0 + e = 2\)

\(e = 2\)

Therefore \(d = 0\) and \(e = 2\)