Ask Me Help Desk - Equations to be solved by Elimination and Substitution Process

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)

- Mathematics (https://www.askmehelpdesk.com/forumdisplay.php?f=199)

- - Equations to be solved by Elimination and Substitution Process (https://www.askmehelpdesk.com/showthread.php?t=624549)

Equations to be solved by Elimination and Substitution Process

Word Problem:
- Part of $3000 is invested at 12%, another part at 13%, and the remainder at 14%. The total yearly income from the three investments is $400. The sum of the amounts invested at 12% and 13% equals the amount invested at 14%. Determine how much is invested at each rate.

Equations:
1) x y z=3000
2) .12x .13y .14z=400
3) x y=z

The equations need to be solved by various methods (elimination, substitution).

Since $x+y =z$ and $x+y+z=3000$ , we get
$x+y+z=z+z=2z=3000 \Rightarrow z=1500$
And since $x+y= z= 1500 \Rightarrow x=1500-y$
So
$.12 \cdot x +.13 \cdot y + .14 \cdot z = 400$
$\Rightarrow .12 \cdot (1500 - y) +.13 \cdot y + .14 \cdot 1500 = 400$
$\Rightarrow 180 -.12 \cdot y +.13 \cdot y + 210 = 400$
$\Rightarrow .01 \cdot y + 390 = 400$
$\Rightarrow .01 \cdot y =10$
$\Rightarrow y =1000$
And since $x=1500-y = 1500-1000 = 500$
So we have $x=500$ , $y=1000$ , and $z=1500$ .

I hope this helps.

is this equation answered using elimiation or substitution?

thanks

All times are GMT -7. The time now is 02:47 PM.