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

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

system of equations using substitution. please help I've been stuck on this lesson for weeks.

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Answer 1 · 2018-02-21T00:01:43

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

Explanation:

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

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

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

$x = 11 - y$ $x=11-y$

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

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

$22 - 5 y = 8$ $22-5y=8$

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

$- 5 y = - 14$ $-5y=-14$

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

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

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

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

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

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

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

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

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

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

Answer 2 · 2018-02-21T00:08:09

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

Explanation:

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

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

$x + y = 11$ $x+y=11$

Subtract $y$ $y$ from both sides

$x + y - y = 11 - y$ $x+y-y=11-y$

$x = 11 - y$ $x=11-y$

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

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

$2 (- y + 11) - 3 y = 8$ $2(-y+11)-3y=8$

Use the distributive property

$(2) (- y) + (2) (11) - 3 y = 8$ $(2)(-y)+(2)(11)-3y=8$

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

$- 5 y + 22 = 8$ $-5y + 22 = 8$

$- 5 y = 8 - 22$ $-5y = 8-22$

#-5y= -14

Divide both sides by $- 5$ $-5$

$\frac{- 5 y}{- 5} = \frac{- 14}{- 5}$ $(-5y)/-5=(-14)/-5$

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

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

$x = - y + 11$ $x=-y+11$

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

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

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

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

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

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

Answer 3 · 2018-02-21T00:17:08

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

Explanation:

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

$x + y = 11 \Rightarrow y = 11 - x$ $x+y=11 => y=11-x$

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

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

$2 x - 3 (11 - x) = 8$ $2x - 3(11-x) = 8$

$2 x - 33 + 3 x = 8$ $2x - 33 + 3x = 8$

$5 x - 33 = 8$ $5x - 33 = 8$

$5 x = 41$ $5x = 41$

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

$x = 8.2$ $x = 8.2$

So

$\Rightarrow y = 11 - (\frac{41}{5})$ $=> y = 11 - (41/5)$

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

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

Or

$y = 11 - (8.2)$ $y = 11 - (8.2)$

$y = 2.8$ $y = 2.8$

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

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

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

Your solution is correct.

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

system of equations using substitution. please help I've been stuck on this lesson for weeks.

3 Answers

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