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$={:[x-y=3],[x^(2)+y^(2)=41]:} substitution mernod )^(2)+y^(2)=9$

$=\begin{array}{c} x-y=3 \\ x^{2}+y^{2}=41 \end{array}$ $\newline$ substitution mernod $\newline$ $)^{2}+y^{2}=9$

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Composition of linear and quadratic functions: find a value

Full solution

Q.

=\begin{array}{c} x-y=3 \\ x^{2}+y^{2}=41 \end{array}

\newline

substitution mernod

\newline

)^{2}+y^{2}=9

Solve for x: Step $1$ : Solve the first equation for x. $\newline$ Given: $x - y = 3$ $\newline$ Solve for x: $x = y + 3$
Substitute $x$ : Step $2$ : Substitute $x$ in the second equation. $\newline$ Given: $x^2 + y^2 = 41$ $\newline$ Substitute $x = y + 3$ into the equation: $\newline$ $(y + 3)^2 + y^2 = 41$
Expand and simplify: Step $3$ : Expand and simplify the equation. $\newline$ $(y + 3)^2 + y^2 = 41$ $\newline$ $y^2 + 6y + 9 + y^2 = 41$ $\newline$ $2y^2 + 6y + 9 = 41$
Solve the quadratic: Step $4$ : Solve the quadratic equation. $\newline$ $2y^2 + 6y + 9 = 41$ $\newline$ $2y^2 + 6y - 32 = 0$ $\newline$ Divide by $2$ : $\newline$ $y^2 + 3y - 16 = 0$
Factorize the quadratic: Step $5$ : Factorize the quadratic equation. $\newline$ $y^2 + 3y - 16 = 0$ $\newline$ $(y + 8)(y - 2) = 0$
Find values of y: Step $6$ : Find the values of y. $\newline$ $y + 8 = 0$ or $y - 2 = 0$ $\newline$ $y = -8$ or $y = 2$
Substitute back to find x: Step $7$ : Substitute y back to find x. $\newline$ Using $x = y + 3$ : $\newline$ For $y = -8$ : $x = -8 + 3 = -5$ $\newline$ For $y = 2$ : $x = 2 + 3 = 5$

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