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Sagot :
We can use a system of linear equations to solve this. I will use the variable P for phones and A for accessories.
Our first equation is: 18P + 21A = 204. This is because she told 18 phones and 21 accessories yesterday which got her $204.
Our next equation is: 6P + 14A = 82. This is because she sold 6 phones and 14 accessories today which got her $82.
We can use elimination to solve this system. We can multiply the equation 6P + 14A = 82 by -3. This is so we can avoid sign errors if we multiply by 3. Our new equation is -18P - 42A = -246.
Next, we add that equation to 18P + 21A = 204. This gives us -21A = -42. Once we solve for A, by dividing by -21, we get A = 2.
Next, we can plug A into our equation. I will plug it into the second one. 6P + 14(2) = 82. This gives us 6P + 28 = 82. Once we get 6P alone, we get 6P = 54. Solving for P gives us 9.
One phone costs $9, and an accessory costs $2.
Our first equation is: 18P + 21A = 204. This is because she told 18 phones and 21 accessories yesterday which got her $204.
Our next equation is: 6P + 14A = 82. This is because she sold 6 phones and 14 accessories today which got her $82.
We can use elimination to solve this system. We can multiply the equation 6P + 14A = 82 by -3. This is so we can avoid sign errors if we multiply by 3. Our new equation is -18P - 42A = -246.
Next, we add that equation to 18P + 21A = 204. This gives us -21A = -42. Once we solve for A, by dividing by -21, we get A = 2.
Next, we can plug A into our equation. I will plug it into the second one. 6P + 14(2) = 82. This gives us 6P + 28 = 82. Once we get 6P alone, we get 6P = 54. Solving for P gives us 9.
One phone costs $9, and an accessory costs $2.
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