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Alex777 [14]
3 years ago
7

Under what circumstances does the system of equations Qx+Ry=S and Y=Tx+S have infinitely many solutions?

Mathematics
1 answer:
Anettt [7]3 years ago
4 0
From these -Tx+y=S. If -T=Q/R, then y=-Qx/R+S, so Ry=-Qx+RS, Qx+Ry=RS=S.
If R is not equal to 1, or S is non-zero, the equations are inconsistent, so there would be no solutions.
If R=1 there are an infinite number of solutions given by Qx+y=S, or y=S-Qx or y=S+Tx.
If S=0, Qx+Ry=0 or y=-Qx/R or y=Tx.
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Explanation:\begin{gathered} \text{Given:} \\ \frac{9\text{ }\times10^9}{4.5\text{ }\times10^1} \end{gathered}

let's break down the expression to get the final result:

\begin{gathered} \frac{9\text{ }\times10^9}{4.5\text{ }\times10^1}\text{ = }\frac{9}{4.5}\times\text{ }\frac{10^9}{10^1} \\ \frac{9}{4.5}\text{ = }\frac{9\text{ }\times\text{ 10}}{4.5\text{ }\times\text{ 10}}\text{ = }\frac{90}{45} \\ \frac{9}{4.5}\text{ = }2 \\  \\ \frac{10^9}{10^1}\colon\text{ when we divide exponents with same base, } \\ we\text{ take one of the base and combine the exponents by subtracting them:} \\ \frac{10^9}{10^1}=10^{9-1} \\ \frac{10^9}{10^1}=10^8 \end{gathered}\begin{gathered} \frac{9}{4.5}\times\text{ }\frac{10^9}{10^1}\text{ = 2 }\times10^8 \\  \\ \text{when dividing decimals, }the\text{ least number of }significant\text{ }figures\text{ in the problem}, \\ \text{ }\det ermines\text{ the significant figures in the answer} \\ 9\text{ = 1 significant, 4.5 = 2 significant} \\ \text{The least significant is 1} \\  \\ \frac{9\text{ }\times10^9}{4.5\text{ }\times10^1}\text{ = 2 }\times10^8 \end{gathered}

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