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

Ch. 5 - Systems of Equations and Inequalities

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Blitzer 8th Edition

Ch. 5 - Systems of Equations and Inequalities

Problema 23

Capitolo 6, Problema 23

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}$

Guida verificata passo dopo passo

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 $y^{2}$.

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