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4x^(2)+x-29=y

x-4=y
If
(x,y) is a solution to the system of equations shown and
x > 0, what is the value of
x ?

$4 x^{2}+x-29=y$ $\newline$ $x-4=y$ $\newline$ If $(x, y)$ is a solution to the system of equations shown and x>0 , what is the value of $x$ ?

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Solve linear equations: mixed review

Full solution

Q.

4 x^{2}+x-29=y

\newline

x-4=y

\newline

If

(x, y)

is a solution to the system of equations shown and

x>0

, what is the value of

x

?

Set up equations: Set up the system of equations. $\newline$ We have two equations: $\newline$ $1$ ) $4x^2 + x - 29 = y$ $\newline$ $2$ ) $x - 4 = y$ $\newline$ Since both expressions are equal to $y$ , we can set them equal to each other.
Equate expressions for y: Equate the two expressions for y. $\newline$ $4x^2 + x - 29 = x - 4$ $\newline$ Now we will solve for $x$ .
Isolate x terms: Subtract $x$ from both sides to start isolating $x$ terms on one side. $\newline$ $4x^2 + x - x - 29 = x - x - 4$ $\newline$ This simplifies to: $\newline$ $4x^2 - 29 = -4$
Isolate quadratic term: Add $29$ to both sides to isolate the quadratic term. $\newline$ $4x^2 - 29 + 29 = -4 + 29$ $\newline$ This simplifies to: $\newline$ $4x^2 = 25$
Solve for $x^2$ : Divide both sides by $4$ to solve for $x^2$ . $\newline$ $\frac{4x^2}{4} = \frac{25}{4}$ $\newline$ This simplifies to: $\newline$ $x^2 = 6.25$
Take square root: Take the square root of both sides to solve for $x$ . $\newline$ Since x > 0, we only consider the positive square root. $\newline$ $x = \sqrt{6.25}$ $\newline$ $x = 2.5$

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