Question #bbb03

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Answer 1 · 2017-12-15T04:37:39

You solve this as a systems of equations.

Start by multiplying the entire equation of $y=4x-7$ by $7$ and the entire equation of $y=-7x+15$ by $4$ . Now you have:

$7y=28x-49" "$ and $" "4y=-28x+60$

Line them up vertically:

$7y=28x-49$
$4y=-28x+60$

and subtract them, so the $x$ 's cancel out. You now should have

$3y=11$

Divide both sides by $3$ to get $y$ by itself, and you get

$y = 11/3.$

Now, plug in $11/3$ for $y$ into the original equation and solve for $x$ :

$11/3=4x-7" "$ (add $7$ to both sides to cancel it)

$32/3=4x " "$ (divide both sides by $4$ to cancel it)

$8/3 = x$

BAM! You're done!

Answer 2 · 2017-12-15T12:15:53

$x=2 and y=1$

This is the best form to have when solving a system of equations,!

They both show $y$ in terms of $x$ .

We know that $color(red)(y = y)$

So, the expressions that are both equal to $y$ must be equal!

$color(white)(xxxx)color(red)(y = y)$

$color(red)(4x-7 = -7x+15)" "larr$ solve for $x$

$4x +7x = 15+7$

$11x = 22$

$x = 2$

Substitute $2$ for $x$ in both equations to check that you get the same answer.

$y = 4x -7" "and" "y = -7x+15$

$y = 4(2) -7" "and" "y = -7(2)+15$

$y = 8-7" "and " "y = -14+15$

In both cases $y=1$