HSA.REI.C.6 math practice. Learn by doing.

Add the equations: $3x = 12$, so $x = 4$. From $x - y = 3$, $y = 1$. Then $x + 3y = 4 + 3 = 7$.

Substitute $y = -x + 8$ into the second equation: $4x - x + 8 = 14$, so $3x = 6$ and $x = 2$. Then $y = 6$, and $x + y = 8$.

Let $x$ be the first integer and $y$ the second. Then $xy = 260$ and $x = 2y + 6$. From the product, $y = \dfrac{260}{x}$. Substitute into the second equation: $x = 2\left(\dfrac{260}{x}\right) + 6$. Multiply through by $x$ to clear the denominator: $x^{2} = 520 + 6x$. Rearrange: $x^{2} - 6x - 520 = 0$. Factor: $(x - 26)(x + 20) = 0$. The positive solution is $x = 26$, so $y = \dfrac{260}{26} = 10$. The smaller integer is $10\textsf{.}$

The parts add to $28$, so $a + b = 28$. Also $a = 3b + 4$. Substitute: $(3b + 4) + b = 28$, so $4b + 4 = 28$, $4b = 24$, and $b = 6$. Then $a = 3(6) + 4 = 22\textsf{.}$

The situation can be written as the system $\begin{cases} 3x + 8y = 67 \\ y = 5 \end{cases}$. Substitute $y = 5$ into the first equation: $3x + 8(5) = 67$, so $3x + 40 = 67$. Then $3x = 27$ and $x = 9$.

Solve using substitution.
$\begin{aligned}&y=\textcolor{red}{x+3} \\ &x\textcolor{red}{+y}\,=9 \\ & \\ &x+\textcolor{red}{(x+3)}=9 \\ &2x+3=9 \\ &2x=6 \\ &x=3 \\ & \\ &y=\textcolor{blue}{x}+3 \\ &y=\textcolor{blue}{3}+3 \\ &y=6 \qquad \text{The solution is }(3, 6)\textsf{.}\end{aligned}$

Substitute $x = 6$ into the second equation: $5(6) + 2y = 52$, so $30 + 2y = 52$, $2y = 22$, and $y = 11$.

Substitute $y = 7$ into $y = 2x - 5$: $7 = 2x - 5$, so $12 = 2x$ and $x = 6$. The solution is $(6, 7)$.