How do you solve $9 x + 6 y = 3$ $9x+6y=3$ and $- 3 x + y = 14$ $-3x+y=14$ using substitution?

Question

How do you solve $9 x + 6 y = 3$ $9x+6y=3$ and $- 3 x + y = 14$ $-3x+y=14$ using substitution?

1 Answer

Impact of this question

2670 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-01T00:20:09

First, you take 1 equation and you make it so that the $y$ $y$ is the main thing like $y =$ $y =$ something... Then you take that and you substitute it into your other equation to get $x =$ $x =$ something...You then take that and you substitute it into your first equation and done.

Explanation:

NOTE:
color $g o l d$ $color(gold)(gold$ and the bolded text is the main stuff
Other writing is explanations

So, to start you got these 2 equations...
$9 x + 6 y = 3$ $color(gold)(9x + 6y = 3$
$- 3 x + y = 14$ $color(gold)(-3x + y = 14$

Let's call equation 1 A and equation 2 B. So...
$9 x + 6 y = 3$ $9x + 6y = 3$ : $A$ $color(red)(A$
$- 3 x + y = 14$ $-3x + y = 14$ : $B$ $color(blue)(B$

So to start substituting, you first take $B$ $color(blue)(B$ because it has the lowest number of ' $y$ $y$ 's...
$- 3 x + y = 14$ $-3x +y = 14$ $B$ $color(blue)(B$
$9 x + 6 y = 3$ $9x + 6y = 3$ $A$ $color(red)(A$
Equation $B$ $color(blue)(B$ has the lowest number of $y$ $y$ .

Now, you take equation $B$ $color(blue)(B$ and you twist and turn it to make the $y$ $y$ in the equation the main letter. Like so...

$- 3 x + y = 14$ $color(gold)(-3x + y = 14$ | This is your starting equation
$y = 14 + 3 x$ $y = 14 + 3x$ | you move over the ' $- 3 x$ $-3x$ ' to the other side of the ' $=$ $=$ ' sign. Since it was a 'negative' number before, going over to the other side makes ' $- 3 x$ $-3x$ ' a positive number, ' $3 x$ $3x$ '.
$y = 14 + 3 x$ $color(gold)(y = 14 + 3x$ | now ' $y$ $y$ ' is the main letter.

Now you have this, you get your other equation (equation $A$ $color(red)(A$ ) and you 'combine' it with your new equation $B$ $color(blue)(B$ . Like so...

$9 x + 6 y = 3$ $9x + 6y = 3$ : $A$ $color(red)(A$
$y = 14 + 3 x$ $y = 14 + 3x$ : NEW $B$ $color(blue)(B$

You substitute your NEW $B$ $color(blue)(B$ into $A$ $color(red)(A$ ...
$9 x + 6 (14 + 3 x) = 3$ $9x+ 6color(gold)((14 + 3x)) = 3$ | In the NEW $B$ $color(blue)(B$ , it says the $y =$ $y =$ 'that'. Therefore, you can make the $y$ $y$ in $A$ $color(red)(A$ ( $6 y$ $6y$ ) also equal 'that'.

Now you just have to do some switch-a-roos and stuff to get your answers: x = something and y = something, like this...

$9 x + 6 (14 + 3 x) = 3$ $color(gold)(9x + 6(14 + 3x) = 3$ | what you got now
$9 x + 84 + 18 x = 3$ $9x + 84 + 18x = 3$ | expanded the brackets
$27 x + 84 = 3$ $27x + 84 = 3$ | added the similar numbers
$27 x = 3 - 84$ $27x = 3 - 84$ | moved $84$ $84$ to the other side
$27 x = - 81$ $27x = -81$
$x = - \frac{81}{27}$ $x = -81/27$ | moved $27$ $27$ to the other side.
$x = - 3$ $color(gold)(x = -3$ | your first answer.

Now...
$- 3 x + y = 14$ $color(gold)(-3x + y = 14$ : $B$ $color(blue)(B$ | The old equation $B$ $color(blue)(B$
Since $x = - 3$ $x = -3$ so...
$- 3 (- 3) + y = 14$ $-3(-3) + y = 14$ | substituting $x = - 3$ $x = -3$
$9 + y = 14$ $9 + y = 14$
$y = 14 - 9$ $y = 14 - 9$ | moving 9 to the other side
$y = 5$ $color(gold)(y = 5$ | your second answer.

That's how you do it.

Science

Math

Humanities

... and beyond

How do you solve $9 x + 6 y = 3$ $9x+6y=3$ and $- 3 x + y = 14$ $-3x+y=14$ using substitution?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 9x+6y=39x+6y=39x+6y=3 and -3x+y=14−3x+y=14-3x+y=14 using substitution?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $9 x + 6 y = 3$ $9x+6y=3$ and $- 3 x + y = 14$ $-3x+y=14$ using substitution?