# Give the solution to the system of equations. x^2+y^2=5, y=3x-5

Give the solution to the system of equations.
${x}^{2}+{y}^{2}=5,y=3x-5$
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Step 1
Give the solution to the system of equations.
${x}^{2}+{y}^{2}=5,y=3x-5$
Step 2
Given
${x}^{2}+{y}^{2}=5$...(i)
y=3x-5...(ii)
Step 3
Substituting (ii) in (i) and solving we get
${x}^{2}+{y}^{2}=5$...(i)
y=3x-5...(ii)
${x}^{2}+{\left(3x-5\right)}^{2}=5$
$⇒{x}^{2}+{\left(3x\right)}^{2}+{\left(5\right)}^{2}-2\left(3x\right)\left(5\right)=5$
$⇒{x}^{2}+9{x}^{2}+25-30x=5$
$⇒{x}^{2}+9{x}^{2}+25-30x-5=5-5$
$⇒{x}^{2}+9{x}^{2}+25-30x-5=0$
$⇒10{x}^{2}-30x+20=0$
$⇒10\left({x}^{2}-3x+2\right)=0$
$⇒{x}^{2}-3x+2=0$
$⇒{x}^{2}-2x-x+2=0$
$⇒x\left(x-2\right)-1\left(x-2\right)=0$
$⇒\left(x-2\right)\left(x-1\right)=0$
$⇒x-2=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x-1=0$
$⇒x=2\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=1$
Step 4
Plugging x=1 and x=2 in (ii) we get
y=3x-5
at x = 1
y=3(1)-5
$⇒y=3-5$
$⇒y=-2$
at x = 2
y=3(2)-5
$⇒y=6-5$
$⇒y=1$
Hence(1,-2) and (2,1)
Jeffrey Jordon