Question 1066108
(1) {{{ y = 2x + 5 }}}
(2) {{{ y = 2x^2 - x + 3 }}}
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Substitute (1) into (2)
(2) {{{ 2x + 5 = 2*x^2 - x + 3 }}}
(2) {{{ 2x^2 - 3x - 2 = 0 }}}
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Use the quadratic formula
{{{ x = ( -b +- sqrt( b^2-4*a*c )) / (2*a) }}} 
{{{ a = 2 }}}
{{{ b = -3 }}}
{{{ c = -2 }}}
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{{{ x = ( -(-3) +- sqrt( (-3)^2-4*2*(-2) )) / (2*2) }}} 
{{{ x = ( 3 +- sqrt( 9 + 16 )) / 4 }}} 
{{{ x = ( 3 +- sqrt( 25 ) ) / 4 }}}
{{{ x = ( 3 + 5 ) / 4 }}}
{{{ x = 2 }}}
and
{{{ x = ( 3 - 5 ) / 4 }}}
{{{ x = -2/4 }}}
{{{ x = -1/2 }}}
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Plug the x's back into (1) or (2) to get y's
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(1) {{{ y = 2x + 5 }}}
(1) {{{ y = 2*2 + 5 }}}
(1) {{{ y = 9 }}}
and
(1) {{{ y = 2x + 5 }}}
(1) {{{ y = 2*(-1/2) + 5 }}}
(1) {{{ y = -1 + 5 }}}
(1) {{{ y = 4 }}}
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The intersections are:
( 2, 9 ) and ( -1/2, 4 )
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check:
(2) {{{ y = 2x^2 - x + 3 }}}
(2) {{{ y = 2*2^2 - 2 + 3 }}}
(2) {{{ y = 8 - 2 + 3 }}}
(2) {{{ y = 9 }}}
OK
You can check ( -1/2, 4 )
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Here's the plot:
{{{ graph( 400, 400, -6, 6, -6, 10, 2x + 5, 2x^2 - x + 3 ) }}}