These simultaneous equation solve them by both elimination and substitution method

Here are the solutions to the simultaneous equations using both elimination and substitution methods.

1. Equations:

$$3x - y = 8 \quad (1)$$ $$5x + y = 13 \quad (2)$$

• Elimination Method:

  Step 1: Add equation (1) and equation (2) to eliminate $y$.

  $$(3x - y) + (5x + y) = 8 + 13$$ $$8x = 21$$ $$x = \frac{21}{8}$$

  Step 2: Substitute the value of $x$ into equation (1).

  $$3\left(\frac{21}{8}\right) - y = 8$$ $$\frac{63}{8} - y = 8$$ $$y = \frac{63}{8} - \frac{64}{8}$$ $$y = -\frac{1}{8}$$

  The solution is $x = \boxed{\frac{21}{8}}$, $y = \boxed{-\frac{1}{8}}$.

• Substitution Method:

  Step 1: From equation (1), express $y$ in terms of $x$.

  $$y = 3x - 8 \quad (3)$$

  Step 2: Substitute equation (3) into equation (2).

  $$5x + (3x - 8) = 13$$ $$8x - 8 = 13$$ $$8x = 21$$ $$x = \frac{21}{8}$$

  Step 3: Substitute the value of $x$ back into equation (3).

  $$y = 3\left(\frac{21}{8}\right) - 8$$ $$y = \frac{63}{8} - \frac{64}{8}$$ $$y = -\frac{1}{8}$$

  The solution is $x = \boxed{\frac{21}{8}}$, $y = \boxed{-\frac{1}{8}}$.

2. Equations:

$$6x + 2y = 11 \quad (1)$$ $$2x + 3y = 1 \quad (2)$$

• Elimination Method:

  Step 1: Multiply equation (2) by 3 to make the coefficients of $x$ equal.

  $$3(2x + 3y) = 3(1)$$ $$6x + 9y = 3 \quad (3)$$

  Step 2: Subtract equation (3) from equation (1).

  $$(6x + 2y) - (6x + 9y) = 11 - 3$$ $$-7y = 8$$ $$$$