Solve Systems of Equations by Substitution: exemplo 5.17

Problema e solução em inglês, como no livro original.

Solve the system by substitution. ${\begin{matrix} y = −2 x + 5 \\ y = \frac{1}{2} x \end{matrix}$

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Sua vez

O Try It do livro logo depois deste exemplo: mesma ideia, outros números. Só a resposta é dada.

Solve the system by substitution. ${\begin{matrix} y = 3 x - 16 \\ y = \frac{1}{3} x \end{matrix}$

Como foi?

Substitute $\frac{1}{2} x$ for y in the first equation.
Replace the y with $\frac{1}{2} x .$
Solve the resulting equation. Start by clearing the fraction.
Solve for x.

Substitute x = 2 into y = $\frac{1}{2} x$ to find y.
The ordered pair is (2,1).
Check the ordered pair in both equations: $\begin{matrix} \begin{matrix} y & = & \frac{1}{2} x \\ 1 & \overset{?}{=} & \frac{1}{2} \cdot 2 \\ 1 & = & 1 ✓ \end{matrix} & \begin{matrix} y & = & - 2 x + 5 \\ 1 & \overset{?}{=} & - 2 \cdot 2 + 5 \\ 1 & = & - 4 + 5 \\ 1 & = & 1 ✓ \end{matrix} \end{matrix}$
	The solution is (2,1).

Substitute $\frac{1}{2} x$ for y in the first equation.
Replace the y with $\frac{1}{2} x .$
Solve the resulting equation. Start by clearing the fraction.
Solve for x.

Substitute x = 2 into y = $\frac{1}{2} x$ to find y.
The ordered pair is (2,1).
Check the ordered pair in both equations: $\begin{matrix} \begin{matrix} y & = & \frac{1}{2} x \\ 1 & \overset{?}{=} & \frac{1}{2} \cdot 2 \\ 1 & = & 1 ✓ \end{matrix} & \begin{matrix} y & = & - 2 x + 5 \\ 1 & \overset{?}{=} & - 2 \cdot 2 + 5 \\ 1 & = & - 4 + 5 \\ 1 & = & 1 ✓ \end{matrix} \end{matrix}$
	The solution is (2,1).