Question 179292
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Let Joan's age be represented by <i>x</i>.  Then four times Joan's age would be 4<i>x</i>.


Let Jim's age be represented by <i>y</i>.  Then three times Jim's age would be 3<i>y</i>.


Four times Joan's age (4<i>x</i>), the sum of (+), three times Jim's age (3<i>y</i>) is (=) 47.  So:  *[tex \Large 4x + 3y = 47]


Jim is (<i>y</i> =) one year less than twice as old as Joan (2<i>x</i> - 1). So: *[tex \Large y = 2x - 1]


Substitute 2<i>x</i> - 1 for <i>y</i> in *[tex \Large 4x + 3y = 47]


*[tex \text {         } \math \Large 4x + 3(2x - 1) = 47]


*[tex \text {         } \math \Large 4x + 6x - 3 = 47]


*[tex \text {         } \math \Large 10x = 50]


*[tex \text {         } \math \Large x = 5]


*[tex \text {         } \math \Large y = 2(5) - 1 = 9]


Check:


*[tex \text {         } \math \Large 4(5) + 3(9) = 20 + 27 = 47] Answer checks.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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