Consider the following system of equations:
$x^2+y^2$x2+y2 | $=$= | $4$4 |
$3x+4y$3x+4y | $=$= | $0$0 |
By filling in the missing values, verify that the points of intersection on the graphs are solutions of the corresponding system of equations.
First, test the point $\left(\frac{8}{5},-\frac{6}{5}\right)$(85,−65).
$x^2+y^2=4$x2+y2=4 |
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$LHS$LHS | $=$= | $\left(\editable{}\right)^2+\left(\editable{}\right)^2$()2+()2 | $RHS$RHS | $=$= | $\editable{}$ | |
$=$= | $\frac{64}{25}+\editable{}$6425+ | |||||
$=$= | $\frac{\editable{}}{25}$25 | |||||
$=$= | $\editable{}$ | |||||
$3x+4y=0$3x+4y=0 |
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$LHS$LHS | $=$= | $3\times\editable{}+4\times\editable{}$3×+4× | $RHS$RHS | $=$= | $\editable{}$ | |
$=$= | $\editable{}+\left(-\frac{24}{5}\right)$+(−245) | |||||
$=$= | $\editable{}$ |
Now test the point $\left(-\frac{8}{5},\frac{6}{5}\right)$(−85,65).
$x^2+y^2=4$x2+y2=4 |
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$LHS$LHS | $=$= | $\left(\frac{\editable{}}{\editable{}}\right)^2+\left(\frac{\editable{}}{\editable{}}\right)^2$()2+()2 | $RHS$RHS | $=$= | $\editable{}$ | |
$=$= | $\editable{}+\frac{36}{25}$+3625 | |||||
$=$= | $\frac{\editable{}}{25}$25 | |||||
$=$= | $\editable{}$ | |||||
$3x+4y=0$3x+4y=0 |
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$LHS$LHS | $=$= | $3\times\editable{}+4\times\editable{}$3×+4× | $RHS$RHS | $=$= | $\editable{}$ | |
$=$= | $-\frac{24}{5}+\editable{}$−245+ | |||||
$=$= | $\editable{}$ |
Consider the system of equations.
$x^2+y^2=10$x2+y2=10
$x-y=2$x−y=2
Consider the system of equations.
$x^2+y^2=2$x2+y2=2
$-3x+2y=5$−3x+2y=5
A system of two equations in two variables whose graphs are a parabola and a circle can have four real ordered-pair solutions.