How do you solve the simultaneous equations 4x + 6y = 16 and x + 2y = 5?

Jul 26, 2015

I found:
$x = 1$
$y = 2$

Explanation:

You can try multiplying the second equation by $- 3$ and then add the two equations ogether (in column):
{4x+6y=16
{color(red)(-3x-6y=-15 add:
$x + 0 = 1$
$x = 1$
Substitute into the first:
$4 \cdot 1 + 6 y = 16$
$6 y = 12$
$y = \frac{12}{6} = 2$

Jul 26, 2015

You can use the substitution method to solve this system of equations.

Explanation:

Your two equations look like this

$\left\{\begin{matrix}4 x + 6 y = 16 \\ x + 2 y = 5\end{matrix}\right.$

The first thing you need to do is use of the equations to solve for one of the two variables, $x$ or $y$, then use this value into the other equation.

For simplicity, use the second equation to solve for $x$ by adding $- 2 y$ to both sides of the equation

$x + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 y}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 y}}} = 5 - 2 y$

$x = 5 - 2 y$

Now plug this value for $x$ in your first equation and solve for $y$

$4 \cdot \left(5 - 2 y\right) + 6 y = 16$

$20 - 8 y + 6 y = 16$

$20 - 2 y = 16$

Isolate $y$ on one side of the equation

$- 2 y + \textcolor{red}{\cancel{\textcolor{b l a c k}{20}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{20}}} = 16 - 20$

$- 2 y = - 4 \implies y = \frac{- 4}{- 2} = 2$

Now take this value and use it to determine $x$

$x = 5 - 2 y$

$x = 5 - 2 \cdot 2 = 1$

The two solutions to your system are

$\left\{\begin{matrix}x = \textcolor{g r e e n}{1} \\ y = \textcolor{g r e e n}{2}\end{matrix}\right.$