Question

my question is
$eq {}^{n} 1 \: \: 0.8x + 0.5y = 0.5 \\ eq {}^{n} 2 \: \: 0.5x + 0.7y = 0.74$
the formula is
$x = \frac{b1c2 - b2c1}{a1b2 -a2b1 } \\ y = \frac{a2c1 - a1c2}{a1b2 - a2b1}$

Answer 1

I hv never seen this before

Answer 2

Answer:

$Analysing the question,</p><p></p><p>Given:</p><p></p><p>Simple interest = ₹ 23940</p><p></p><p>Principal = ₹ 28500</p><p></p><p>Time period = 7 years</p><p></p><p>\begin{gathered}\\\end{gathered}</p><p></p><p>To find:</p><p></p><p>Rate of interest per annum = ?</p><p></p><p>\begin{gathered}\\\end{gathered}</p><p></p><p>Formula used:</p><p></p><p>\begin{gathered} \maltese \: \: \boxed { \bf{SI = \dfrac{PTR}{100}}} \\ \\ \end{gathered}✠SI=100PTR</p><p></p><p>where:</p><p></p><p>SI is the Simple interest</p><p></p><p>P is the principal</p><p></p><p>T is the time period</p><p></p><p>R is the rate of interest</p><p></p><p>\begin{gathered}\\\end{gathered}</p><p></p><p>Solution:</p><p></p><p>Applying the above formula and substituting the given values,</p><p></p><p>\begin{gathered}\hookrightarrow \: \: \sf{23940 = \dfrac{28500 \times 7 \times R}{100} } \\ \\ \end{gathered}↪23940=10028500×7×R</p><p></p><p>\begin{gathered} \dashrightarrow \: \: \sf{23940 = \dfrac{285 \cancel{00} \times 7 \times R}{ \cancel{100}} } \\ \\ \end{gathered}⇢23940=10028500×7×R</p><p></p><p>\begin{gathered} \dashrightarrow \: \: \sf{285 \times 7 \times R = 23940} \\ \\ \end{gathered}⇢285×7×R=23940</p><p></p><p>\begin{gathered} \dashrightarrow \: \: \sf{1995 \times R = 23940} \\ \\ \end{gathered}⇢1995×R=23940</p><p></p><p>\begin{gathered} \dashrightarrow \: \: \sf{R = \dfrac{23940}{1995} } \\ \\ \end{gathered}⇢R=199523940</p><p></p><p>$

Step-by-step explanation:

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