3.2x-9.6=-.4x+1.2
3.2x=-.4x+10.8
3.6x=10.8
X= 3
Answer: tea = 15 rupees per kg
sugar= 3 rupees per kg
Step-by-step explanation:
Hi, to answer this question we have to write a system of equations with the information given:
<em>"Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997":
</em>
2 t + 3 s =39 (a)
Where:
- t= price of 1 kg of tea
- s = price of 1 kg of sugar
<em>"in march 1997 the price of the tea increased by 25% (1.25)and the price of the sugar increased by 20%(1.20) and the same quantity of tea and sugar cost rupees 48.30.
"</em>
2(t1.25)+3(s1.2) = 48.30 (b)
- <em>Solving for t in (b)
</em>
2t =39-3s
t = (39 -3s)/2
t = 19.5-1.5s
- <em>Replacing the value of t in (b)
</em>
2 x ((19.5-1.5s)1.25)+ 3 ( 1.2s) =48.30
2x ( 24.375 -1.875s) +3.6s =48.30
48.75 -3.75s+3.6s= 48.30
48.75-48.30 = 3.75s-3.6s
0.45= 0.15s
0.45/0.15 =s
3 =s
- <em>Replacing the value of s in (a)
</em>
2 t + 3 (3) =39
2 t + 9 =39
2 t =39 -9
2 t =30
t = 30/2
t= 15
Prices in january:
tea = 15 rupees per kg
sugar= 3 rupees per kg
Feel free to ask for more if needed or if you did not understand something.
So the two possible original temperatures would be either -9 or 9.
1.5x will be the very answer
Answer:
56
Step-by-step explanation:
Strange question to me
