document.write( "Question 470832: Solve the system by the method of substitution.\r
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document.write( "{x^2+y^2=25
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document.write( "{3x-4y=0\r
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document.write( "a. no real solution
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document.write( "b.(3,4),(3,-4)
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document.write( "c.(4,3),(-4,-3)
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document.write( "d.(4,3)
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document.write( "e.(4,3)(-4,3),(-4,-3),(4,-3) \n" );
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Algebra.Com's Answer #322942 by hinaahsan(5) ![]() You can put this solution on YOUR website! x^2 + y^2 =25 --> (1) \n" ); document.write( "whereas, 3x-4y=0 -->(2) \n" ); document.write( " \n" ); document.write( "using equation 2 \n" ); document.write( "3x=4y \n" ); document.write( "x=4y/3 -->(3) \n" ); document.write( " using value of x from equation (3) and subsituting in equation (1)\r \n" ); document.write( "\n" ); document.write( "(4y/3)^2 + y^2 =25 \n" ); document.write( "16y^2/9 +y^2 =25\r \n" ); document.write( "\n" ); document.write( "(16y^2/9 +9y^2/9) = 25\r \n" ); document.write( "\n" ); document.write( "25y^2/9 =25 \n" ); document.write( "y^2/9 =25/25\r \n" ); document.write( "\n" ); document.write( "y^2/9 =1\r \n" ); document.write( "\n" ); document.write( "y^2 = 9\r \n" ); document.write( "\n" ); document.write( "(y)^2 = (3)^2\r \n" ); document.write( "\n" ); document.write( "y = 3 -->(4)\r \n" ); document.write( "\n" ); document.write( "using the value of y from equation (4) in equation (3) to get the value of x\r \n" ); document.write( "\n" ); document.write( "x= 4y/3 \n" ); document.write( "x= 4(3)/3 \n" ); document.write( "x=4\r \n" ); document.write( "\n" ); document.write( "Ans: (x,y)=(4,3) \n" ); document.write( " |