Question 188895
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1. Two numbers wich differ by 3, have a product of 88. Find them


Let x = 1st number
x-3 = 2nd number (differ by 3)
Then,
{{{(x)(x-3)=88}}}
{{{x^2-3x-88=0}}}
{{{(x-11)(x+8)=0}}}
{{{highlight(x=-8)}}} or {{{highlight(x=11)}}}, 1st number
{{{-8-3=highlight(-11)}}} or {{{11-3=highlight(8)}}}, 2nd number


Check, {{{11*8=88}}}, good. Or {{{-11*-8=88}}}, also good.



2. The product of two consecutive odd numbers is 143. Form an algebraic equation and solve it to find the numbers.


Let x= 1st odd number
x+2= 2nd odd number
Then,
{{{(x)(x+2)=143}}}
{{{x^2+2x-143=0}}}
{{{(x-11)(x+13)=0}}}
{{{highlight(x=11)}}} or {{{highlight(x=-13)}}}, 1st odd number
{{{11+2=highlight(13)}}} or {{{-13+2=highlight(-11)}}}, 2nd odd number


Check, {{{11*13=143}}}, good. Or, {{{-11*-13=143}}}, good.


3. One number exceeds another by 7. Their product is 60. Form an equation in algebra and solve it to find the number. 


Let x= one number
x+7 = another number
Then,
{{{(x)(x+7)=60}}}
{{{x^2+7x-60=0}}}
{{{(x+12)(x-5)=0}}}
{{{highlight(x=-12)}}} or {{{highlight(x=5)}}}, one number
{{{-12+7=highlight(-5)}}} or {{{5+7=highlight(12)}}}, another number


Let's check, {{{-12*-5=60}}}, good. Or, {{{5*12=60}}}, good.


I'll leave the last number for you (#4)  for your exercise.
Just follow thru just like the 1st 3 numbers above.


Thank you,
Jojo

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