How do you solve 6x+5y=10 and 7x-3y=24?

Question

How do you solve 6x+5y=10 and 7x-3y=24?

1 Answer

Impact of this question

2518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-12T05:36:24

There are two common ways:

Substitution
Elimination

I can show both ways, but in this case I think elimination would be easier.

$6 x + 5 y = 10$ $6x + 5y = 10$
$7 x - 3 y = 24$ $7x - 3y = 24$

I got:

$(x, y) = (\frac{150}{53}, - \frac{74}{53})$ $(x,y) = (150/53, -74/53)$

SUBSTITUTION

Solve for one variable first.

$6 x = 10 - 5 y$ $6x = 10 - 5y$

$\Rightarrow x = \frac{5}{3} - \frac{5}{6} y$ $=> x = 5/3 - 5/6 y$

Plug it into the second equation to solve for the other variable.

$7 (\frac{5}{3} - \frac{5}{6} y) - 3 y = 24$ $7(5/3 - 5/6 y) - 3y = 24$

$\frac{35}{3} - \frac{35}{6} y - 3 y = 24$ $35/3 - 35/6 y - 3y = 24$

Multiply through by $6$ $6$ to make this look nicer.

$70 - 35 y - 18 y = 144$ $70 - 35y - 18y = 144$

$- 53 y = 144 - 70$ $- 53y= 144 - 70$

$y = \frac{144 - 70}{- 53} = - \frac{74}{53}$ $color(green)(y) = (144 - 70)/(-53) = color(green)(-74/53)$

Therefore, for $x$ $x$ we get:

$x = \frac{5}{3} - \frac{5}{6} (- \frac{74}{53})$ $color(green)(x) = 5/3 - 5/6(-74/53)$

$= \frac{5}{3} + \frac{370}{318} = \frac{5}{3} + \frac{185}{159}$ $= 5/3 + 370/318 = 5/3 + 185/159$

$= \frac{795}{477} + \frac{555}{477} = \frac{1350}{477}$ $= 795/477 + 555/477 = 1350/477$

$= \frac{150}{53}$ $= color(green)(150/53)$

So, apparently, $(x, y) = (\frac{150}{53}, - \frac{74}{53})$ $color(blue)((x,y) = (150/53, -74/53))$ ... Let's check our answer.

$6 (\frac{150}{53}) + 5 (- \frac{74}{53}) ? = 10$ $6(150/53) + 5(-74/53) stackrel(?" ")(=) 10$

$\frac{900}{53} - \frac{370}{53} ? = 10 \Rightarrow \frac{530}{53} = 10$ $900/53 - 370/53 stackrel(?" ")(=) 10 => 530/53 = 10$ $\sqrt{}$ $color(blue)(sqrt"")$

$7 (\frac{150}{53}) - 3 (- \frac{74}{53}) ? = 24$ $7(150/53) - 3(-74/53) stackrel(?" ")(=) 24$

$\frac{1050}{53} + \frac{222}{53} ? = 24 \Rightarrow \frac{1272}{53} = \frac{12 \cdot 106}{53} = 24 \sqrt{}$ $1050/53 + 222/53 stackrel(?" ")(=) 24 => 1272/53 = (12 cdot 106)/53 = 24 color(blue)(sqrt"")$

Yep, it's right! Wow, not nice-looking at all!

ELIMINATION

In this case we would be scaling one of the equations with fractions to eliminate a variable.

$" "3/5(6x + cancel(5y) = 10)$
$+ " "7x - cancel(3y) = 24$
$bar(" "" "" "" "" "" "" "" ")$
$" "18/5x + 7x = 30$

$(18/5 + 35/5)x = 30$

$color(green)(x) = 30/(18/5 + 35/5)$

$= 30/(53/5) = color(green)(150/53)$

And now, plug it into the second equation.

$7(150/53) - 3y = 24$

$7(150/53) - 24 = 3y$

$color(green)(y) = 7/3(150/53) - 8$

$= 7(50/53) - 424/53$

$= 350/53 - 424/53$

$= color(green)(-74/53)$

And as before, we get:

$color(blue)((x,y) = (150/53, -74/53))$

So if you get something ugly like this, it's right!

Science

Math

Humanities

... and beyond

How do you solve 6x+5y=10 and 7x-3y=24?

1 Answer

Related questions

Impact of this question