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3 years ago

Please help with these 2 !! Thank you

Mathematics

1 answer:

MatroZZZ [7]3 years ago

3 0

Because we have been given the slope and y-intercept, we can create an equation in slope-intercept form.

y = mx + b

m = the slope

b = y-intercept.

In the first problem,

m = -4

b = 10

So, our equation looks like: y = -4x + 10

In the second problem,

m = 1/3

b = -6

So, our equation looks like: y = 1/3x - 6

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QUESTION 1

The given system of equation is

$y=x-6...eqn1$

and

$3x+2y=8...eqn2$

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QUESTION 2

The given system of equations is

$y-x=9...eqn1$

and

$10+2x=-2y...eqn2$

We make y the subject in equation (2) to get,

$y=-x-5..eqn3$

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We divide through by $-2$ to get,

$x=-7$

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The given system is

$y=3x-1..eqn1$

and

$x-y=-9...eqn2$

We make $x$ the subject in equation (2) to get,

$x=y-9...eqn3$

We put equation (3) into equation (1) to obtain,

$y=3(y-9)-1$

We expand the bracket to get,

$y=3y-27-1$

Group like terms to get,

$y-3y=-27-1$

We simplify to get;

$-2y=-28$

This implies that,

$y=14$

Therefore the y-coordinate is 14.

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