Given [latex]x = 2t + 1[/latex], and [latex]y = 3x - 4[/latex], find [latex]y[/latex] in terms of [latex]t[/latex].

We already know what [latex]x[/latex] equals, so substitute [latex]2t + 1[/latex] wherever we see [latex]x[/latex] in [latex]y = 3x - 4[/latex]:

[latex]\begin{align} y &= 3(2t + 1) - 4 \\ &= 6t + 3 - 4 \\ &= 6t - 1 \end{align}[/latex]

So [latex]y = 6t - 1[/latex].

Given [latex]x = 3t - 2[/latex], solve for [latex]t[/latex] and substitute into [latex]y = 2t + 1[/latex].

First, solve [latex]x = 3t - 2[/latex] for [latex]t[/latex]:

[latex]\begin{align} x &= 3t - 2 \\ x + 2 &= 3t \\ t &= \dfrac{x + 2}{3} \end{align}[/latex]

Now substitute this expression for [latex]t[/latex] into [latex]y = 2t + 1[/latex]:

[latex]\begin{align} y &= 2\left(\dfrac{x + 2}{3}\right) + 1 \\ &= \dfrac{2(x + 2)}{3} + 1 \\ &= \dfrac{2x + 4}{3} + 1 \\ &= \dfrac{2x + 4}{3} +\dfrac{3}{3} \\ &= \dfrac{2x + 7}{3} \end{align}[/latex]

So [latex]y = \dfrac{2}{3}x + \dfrac{7}{3}[/latex].

This process connects the two equations into one, eliminating [latex]t[/latex].