How do you solve 2x-3y=8 and x+y=11 using substitution?

Feb 21, 2018

$x = 8.2$, $y = \frac{14}{5}$

Explanation:

Our equations are as follows:
$2 x - 3 y = 8$
$x + y = 11$

Since the second equation is much simpler, we can subtract $y$ from both sides to get an value for $x$ in terms of $y$.

Subtracting $y$ from both sides of the second equation, we get:

$x = 11 - y$

We can now substitute this into the first equation:
$2 \left(11 - y\right) - 3 y = 8$, which simplifies to:

$22 - 2 y - 3 y = 8$, which can be simplified more to:

$22 - 5 y = 8$

We can solve for $y$ now, by subtracting $22$ from both sides and dividing both sides by $- 5.$ We get:

$- 5 y = - 14$

$y = \frac{14}{5}$

We can plug this into either equation to solve for $x .$ Let's plug in $\frac{14}{5}$ in the first equation, $2 x - 3 y = 8$:

$2 x - 3 \left(\frac{14}{5}\right) = 8$

$2 x - \frac{42}{5} = 8$ (We can change 2x to $\frac{10}{5} x$ and 8 into $\frac{40}{5}$ to have a common denominator).

$\frac{10}{5} x - \frac{42}{5} = \frac{40}{5}$

Let's add $\frac{42}{5}$ to both sides:

$\frac{10}{5} x = \frac{82}{5}$

We can multiply both sides by the reciprocal of $\frac{10}{5}$ to solve for $x$.

$x = \frac{82}{5} \left(\frac{5}{10}\right)$

The $5 s$ cancel, and we're left with $\frac{82}{10}$ or $x = 8.2 .$

Feb 21, 2018

$x = \frac{41}{5}$ and $y = \frac{14}{5}$

Explanation:

$2 x - 3 y = 8$
$x + y = 11$

We need to solve $x + y = 11$ for $x$

$x + y = 11$

Subtract $y$ from both sides

$x + y - y = 11 - y$

$x = 11 - y$

now substitute $- y + 11$ for $x$ in $2 x - 3 y = 8$

$2 x - 3 y = 8$

$2 \left(- y + 11\right) - 3 y = 8$

Use the distributive property

$\left(2\right) \left(- y\right) + \left(2\right) \left(11\right) - 3 y = 8$

$- 2 y + 22 - 3 y = 8$

$- 5 y + 22 = 8$

$- 5 y = 8 - 22$

-5y= -14

Divide both sides by $- 5$

$\frac{- 5 y}{-} 5 = \frac{- 14}{-} 5$

$y = \frac{14}{5}$

Now substitute $\frac{14}{5}$ for $y$ in $x = - y + 11$

$x = - y + 11$

$x = - \frac{14}{5} + 11$

$x = \frac{- 14}{5} + \frac{11}{1}$

$x = \frac{- 14}{5} + \frac{11}{5}$

$x = \frac{- 14 + 55}{5}$

$x = \frac{41}{5}$

Answer: $x = \frac{41}{5}$ and y$\frac{14}{5}$

Feb 21, 2018

$\left(x , y\right) = \left(\frac{41}{5} , \frac{14}{5}\right) = \left(8.2 , 2.8\right)$

Explanation:

Isolate $y$ in the second equation by subtracting $x$ on both sides:

$x + y = 11 \implies y = 11 - x$

Substitute $y$ in the first equation with the expression $11 - x$ then solve for $x$, then solve for $y$:

$2 x - 3 y = 8$

$2 x - 3 \left(11 - x\right) = 8$

$2 x - 33 + 3 x = 8$

$5 x - 33 = 8$

$5 x = 41$

$x = \frac{41}{5}$

$x = 8.2$

So

$\implies y = 11 - \left(\frac{41}{5}\right)$

$y = \frac{55}{5} - \frac{41}{5}$

$y = \frac{14}{5}$

Or

$y = 11 - \left(8.2\right)$

$y = 2.8$

Check your solution: insert your values in each of the given equations and verify that they satisfy the system:

2(8.2) - 3(2.8) = 8 " " {"true"}

(8.2) + (2.8) = 11 " "{"true"}#