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3 years ago

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Below are two linear equations . 1: y - 2 = 4/5 * (x + 4) 2: 5x + 4y = - 20 a) Are the lines 1 and 2 parallel , perpendicular or

neither ? b) Which line has a steeper slope, 1 or 2?

1 answer:

Olenka [21]3 years ago

5 0

Perpendicular and 2 I think that’s right

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2 years ago

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3 years ago

2х5y-3z-=14<br> x-2y+=-12<br> -x+ Зу-2z =13

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Answer:

x = -34/3, y = 1/3, and z = -35/3

Step-by-step explanation:

2х+5y-3z-=14 <u>let's call this </u><u>equation 1.</u>

x-2y=-12 <u>let's call this </u><u>equation 2.</u>

-x+Зу-2z =13 <u>let's call this </u><u>equation 3.</u>

<u>USING </u><u>EQUATION 1</u><u> AND </u><u>EQUATION 3</u><u>.</u>

2х+5y-3z-=14 (EQUATION 1)

-x+Зу-2z =13 (EQUATION 3)

<u>LET'S GET RID OF </u><u>z</u><u>, BY MAKING THE COEFFICIENT OF </u><u>z</u><u> IN THE TWO EQUATION THE SAME. WE WILL MULTIPLY </u><u>EQUATION 1</u><u> BY </u><u>2</u><u> AND </u><u>EQUATION 3</u><u> BY </u><u>3</u><u>.</u>

2х+5y-3z-=14 (EQUATION 1) * 2

-x+Зу-2z =13 (EQUATION 3) * 3

4х+10y-6z-=28 <u>let's call this </u><u>equation 4.</u>

-3x+9у-6z =39 <u>let's call this </u><u>equation 5.</u>

<u>TO GET RID OF </u><u>z</u><u>, WE WILL HAVE TO SUBTRACT </u><u>EQUATION 5</u><u> FROM </u><u>EQUATION 4</u><u> PLEASE TAKE NOTE OF THE SIGNS (-) (+).</u>

4х+10y-6z-=28 (EQUATION 4)

- (-3x+9у-6z =39) (EQUATION 5)

(4x - 3x) + ((+10y) - (+9y)) + ((-6z) - (-6z)) = (28 - 39)

x + y + 0 = -11

x + y = -11 <u>let's call this </u><u>equation 6.</u>

<u>USING </u><u>EQUATION 2</u><u> AND </u><u>EQUATION 6</u><u>, LET'S FIND </u><u>x</u><u> AND </u><u>y</u><u>.</u>

x-2y=-12 (EQUATION 2)

x + y = -11 (EQUATION 6)

<u>x</u><u> has the same coefficient in both equations, which is </u><u>1</u><u>. Let's get rid of </u><u>x</u><u> so we can find the value of </u><u>y</u><u>.</u>

<u>We will subtract </u><u>equation 6</u><u> from </u><u>equation 2</u><u>. Take note of the signs.</u>

x-2y=-12 (EQUATION 2)

- (x + y = -11) (EQUATION 6)

(x - x) + ((-2y) - (+y)) = ((-12) - (-11))

0 + -3y = -1

-3y = -1

<u>Divide both sides by </u><u>-3</u><u>.</u>

-3y/-3 = -1/-3

y = 1/3

<u>LET'S SUBSTITUTE THE VALUE OF </u><u>y</u><u> INTO</u><u> EQUATION 2</u><u> TO GET THE VALUE OF </u><u>x</u><u>.</u>

x-2y=-12 (EQUATION 2)

x-2(1/3)=-12

x -2/3 = -12

<u>Add </u><u>2/3</u><u> to both sides.</u>

x -2/3 + (2/3) = -12 + (2/3)

x + 0 = -34/3

x = -34/3

<u>LET'S SUBSTITUTE THE VALUES OF </u><u>x</u><u> and </u><u>y</u><u> INTO</u><u> EQUATION 1</u><u> TO GET THE VALUE OF </u><u>z</u><u>.</u>

2х+5y-3z-=14 (EQUATION 1)

2(-34/3)+5(1/3)-3z-=14

-68/3 + 5/3 - 3z = 14

-21 - 3z = 14

<u>Add </u><u>21</u><u> to both sides.</u>

-21 + (21) - 3z = 14 + (21)

0 - 3z = 35

-3z = 35

<u>Divide both sides by </u><u>-3</u><u>.</u>

-3z/-3 = 35/-3

z = -35/3 <em>or</em>

Therefore, x = -34/3, y = 1/3, and z = -35/3

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6 0

3 years ago