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3 years ago

Evaluate (y - 9) + (3 + x), when y = 15 and x = 5

Mathematics

1 answer:

svet-max [94.6K]3 years ago

5 0

Answer:

Step-by-step explanation:

It says 'Evaluate (y - 9) + (3 + x), when y = 15 and x = 5'.

So we'd put 15 in y's place. It'll look like this (15 - 9).

Then we'd put 5 in x's place. It'll look like this (3 + 5).

You subtract (15 - 9) which equals 6. And you'd add (3 +5) which equals 8. You add 6 + 8 together since those are your answer for the parentheses. Your final should be 14. Hopefully this helped!

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Answer:

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<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

$y = 3x - 1$
$3x + y = -7$

To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

$\implies 3x + y = -7$

$\implies3x + (3x - 1) = -7$

$\implies3x + 3x - 1 = -7$

Combine like terms as needed;

$\implies 3x + 3x - 1 = -7$

$\implies 6x - 1 = -7$

Add 1 to both sides of the equation;

$\implies 6x - 1 + 1 = -7 + 1$

$\implies 6x = -6$

Divide 6 to both sides of the equation;

$\implies \dfrac{6x}{6} = \dfrac{-6}{6}$

$\implies x = -1$

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

$\implies y = 3(-1) - 1$

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Therefore, the value(s) of the x-variable and the y-variable are;

$\boxed{x = -1}$ $\boxed{y = -4}$

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

$\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.$

$\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.$

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

$\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.$

$\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.$

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

$\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.$

$\implies 0 = (6x) + (6)$

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This problem is now an algebraic problem. Isolate "x" to determine its value.

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$\implies -1 = x$

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

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$\boxed{x = -1}$ $\boxed{y = -4}$

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>

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Evaluate (y - 9) + (3 + x), when y = 15 and x = 5