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${:[y=(1)/(5)x-9],[5y^(2)+15=7x]:} If (a,b) is a solution to the system of equations shown and b > 0, what is the value of b ?$

$\begin{aligned} y & =\frac{1}{5} x-9 \\ 5 y^{2}+15 & =7 x \end{aligned}$ $\newline$ If $(a, b)$ is a solution to the system of equations shown and $b>0$ , what is the value of $b$ ?

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Math Problems

Grade 6

Multiply two decimals: where does the decimal point go?

Full solution

\begin{aligned} y & =\frac{1}{5} x-9 \\ 5 y^{2}+15 & =7 x \end{aligned}

\newline

(a, b)

is a solution to the system of equations shown and

b>0

, what is the value of

b

Write Equations: Write down the system of equations. $\newline$ We have the following system of equations: $\newline$ $y = \frac{1}{5}x - 9$ $\newline$ $5y^2 + 15 = 7x$
Express x in terms: Express x in terms of y from the first equation. $\newline$ From the first equation, we can solve for x: $\newline$ $y = \frac{1}{5}x - 9$ $\newline$ $y + 9 = \frac{1}{5}x$ $\newline$ $x = 5(y + 9)$
Substitute x into second: Substitute the expression for x from Step $2$ into the second equation. $\newline$ We substitute $x = 5(y + 9)$ into the second equation: $\newline$ $5y^2 + 15 = 7x$ $\newline$ $5y^2 + 15 = 7(5(y + 9))$
Simplify and solve for y: Simplify the equation and solve for y. $\newline$ $5y^2 + 15 = 35(y + 9)$ $\newline$ $5y^2 + 15 = 35y + 315$ $\newline$ $5y^2 - 35y + 15 - 315 = 0$ $\newline$ $5y^2 - 35y - 300 = 0$
Factor quadratic equation: Factor the quadratic equation. $\newline$ We need to factor the quadratic equation $5y^2 - 35y - 300 = 0$ . This can be done by looking for two numbers that multiply to $5 \times -300 = -1500$ and add up to $-35$ . These numbers are $-60$ and + $25$ . $\newline$ $5y^2 - 60y + 25y - 300 = 0$ $\newline$ $5y(y - 12) + 25(y - 12) = 0$ $\newline$ $(5y + 25)(y - 12) = 0$
Solve for y: Solve for y. $\newline$ Setting each factor equal to zero gives us two possible solutions for y: $\newline$ $5y + 25 = 0$ $\newline$ $y = -5$ $\newline$ and $\newline$ $y - 12 = 0$ $\newline$ $y = 12$
Choose positive y value: Choose the value of y that satisfies the condition b > $0$ . $\newline$ Since we are given that b > $0$ and b corresponds to y in our system of equations, we choose the positive value of y: $\newline$ $y = 12$