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Elementary algebra
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==== Substitution method ==== Another way of solving the same system of linear equations is by substitution. : <math>\begin{cases}4x + 2y &= 14 \\ 2x - y &= 1.\end{cases} </math> An equivalent for {{mvar|y}} can be deduced by using one of the two equations. Using the second equation: : <math>2x - y = 1 </math> Subtracting <math>2x</math> from each side of the equation: : <math>\begin{align}2x - 2x - y & = 1 - 2x \\ - y & = 1 - 2x \end{align}</math> and multiplying by β1: : <math> y = 2x - 1. </math> Using this {{mvar|y}} value in the first equation in the original system: : <math>\begin{align}4x + 2(2x - 1) &= 14\\ 4x + 4x - 2 &= 14 \\ 8x - 2 &= 14 \end{align}</math> Adding ''2'' on each side of the equation: : <math>\begin{align}8x - 2 + 2 &= 14 + 2 \\ 8x &= 16 \end{align}</math> which simplifies to : <math>x = 2 </math> Using this value in one of the equations, the same solution as in the previous method is obtained. : <math>\begin{cases} x = 2 \\ y = 3. \end{cases}</math> This is not the only way to solve this specific system; in this case as well, {{mvar|y}} could have been solved before {{mvar|x}}.
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