Question

In Exercises 19–28, solve each system by the addition method. ﻿{3x2+4y2−16=02x2−3y2−5=0\begin{cases}
$$3x^2$$ + $$4y^2 - 16 = 0$$ \
$$2x^2$$ - $$3y^2 - 5 = 0$$
\end{cases}{3x2+4y2−16=02x2−3y2−5=0​

Accepted Answer

First, write down the system of equations clearly:

$$3x^{2}$$ + $$4y^{2} - 16 = 0$

2x$$^{2}$$ - 3y$$^{2}$$ - 5 = 0$$ Rearrange each equation to isolate the constant on the right side:

$$3x^{2}$$ + $$4y^{2} = 16$

2x$$^{2}$$ - 3y$$^{2}$$ = 5 To $$use the addition method, aim to eliminate one variable by making the coefficients of either $$x^{2} or$$ $$y^{2}$$ opposites. Multiply the first equation by 3 and the second equation by 4 to align the coefficients of $$y^{2}$$:

$$3(3x^{2}$$ + $$4y^{2}) = 3(16$$)$$ which gives 9x$$^{2}$$ + 12y$$^{2}$$ = 48$$

$$4(2x^{2}$$ - $$3y^{2}) = 4(5$$)$$ which gives 8x$$^{2}$$ - 12y$$^{2}$$ = 20$$ Add the two new equations to eliminate $$y^{2}$$:

$$(9x^{2}$$ + $$12y^{2}$$) + $$(8x^{2}$$ - $$12y^{2}) = 48 + 20$

This simplifies to 17x$$^{2}$$ = 68$$ Solve for $$x^{2} by $$dividing both sides by 17:

$$x^{2}$$ = \frac{68}{17}$$

Once you find x$$^{2}$$, substitute this value back into one of the original rearranged equations to solve for y$$^{2}$$.

In Exercises 19–28, solve each system by the addition method. $\begin{cases}\)3x^2 + 4y^2 - 16 = 0 \\2x^2 - 3y^2 - 5 = 0\(\end{cases}$

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Key Concepts

Systems of Equations

Addition Method (Elimination Method)

Quadratic Equations in Two Variables

In Exercises 19–28, solve each system by the addition method. {3x2+4y2−16=02x2−3y2−5=0\(\begin{cases}\)3x^2 + 4y^2 - 16 = 0 \\2x^2 - 3y^2 - 5 = 0\(\end{cases}\){3x2+4y2−16=02x2−3y2−5=0