woowheedr

2022-07-06

1.) Given the following equations:

$3x-y=30\phantom{\rule{0ex}{0ex}}5x-3y=10$

What are the values of $x$ and $y$?

$3x-y=30\phantom{\rule{0ex}{0ex}}5x-3y=10$

What are the values of $x$ and $y$?

trantegisis

Beginner2022-07-07Added 20 answers

You have what we call a system of two linear equations in two unknowns.

$\begin{array}{}\text{(1)}& 3x-y& =30\text{(2)}& 5x-3y& =10\end{array}$

There are a number of ways you can solve for the $x$,$y$ values that satisfy both equations. One way you can approach this is by substitution: expressing $y$ in equation (1) as a function of $x$, and "plugging" that function into into "$y$" in equation 2:

$\begin{array}{}\text{(1)}& 3x-y=30\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{30}}\end{array}$

$\begin{array}{rl}\text{(2)}& 5x-3{\mathbf{y}}& =105x-3({\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{30}})& =10\\ 5x-9x+90& =10\\ -4x& =-80\\ \mathbf{x}& =\mathbf{20}\end{array}$

Now, go back to $(1)$ and "plug in" $x=20$ to solve for $y$:

$\begin{array}{rl}(1)\phantom{\rule{1em}{0ex}}x=20,\phantom{\rule{1em}{0ex}}3x-y=30& \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}3(20)-y=30\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}-y=-30\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathbf{y}\mathbf{=}\mathbf{30}\end{array}$

We have a unique solution to the system given by equations $(1)$ and $(2)$, namely, $x=20,\phantom{\rule{thickmathspace}{0ex}}y=30$, and this corresponds to option $(a)$.

$\begin{array}{}\text{(1)}& 3x-y& =30\text{(2)}& 5x-3y& =10\end{array}$

There are a number of ways you can solve for the $x$,$y$ values that satisfy both equations. One way you can approach this is by substitution: expressing $y$ in equation (1) as a function of $x$, and "plugging" that function into into "$y$" in equation 2:

$\begin{array}{}\text{(1)}& 3x-y=30\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{30}}\end{array}$

$\begin{array}{rl}\text{(2)}& 5x-3{\mathbf{y}}& =105x-3({\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{30}})& =10\\ 5x-9x+90& =10\\ -4x& =-80\\ \mathbf{x}& =\mathbf{20}\end{array}$

Now, go back to $(1)$ and "plug in" $x=20$ to solve for $y$:

$\begin{array}{rl}(1)\phantom{\rule{1em}{0ex}}x=20,\phantom{\rule{1em}{0ex}}3x-y=30& \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}3(20)-y=30\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}-y=-30\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathbf{y}\mathbf{=}\mathbf{30}\end{array}$

We have a unique solution to the system given by equations $(1)$ and $(2)$, namely, $x=20,\phantom{\rule{thickmathspace}{0ex}}y=30$, and this corresponds to option $(a)$.

vortoca

Beginner2022-07-08Added 2 answers

Hint:

Just substitute in your equations $x$ and $y$ by the given values and check the answer.

Just substitute in your equations $x$ and $y$ by the given values and check the answer.

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