8.5 Adding and subtracting rational expressions (Page 2/2)

Elementary algebra Page 2 / 2

$\begin{array}{l} \frac{3 x^{2} + 4 x + 5}{(x + 6) (x - 2)} + \frac{2 x^{2} + x + 6}{x^{2} + 4 x - 12} - \frac{x^{2} - 4 x - 6}{x^{2} + 4 x - 12} \\ Factor the denominators to determine if they're the same . \\ \frac{3 x^{2} + 4 x + 5}{(x + 6) (x - 2)} + \frac{2 x^{2} + x + 6}{(x + 6) (x - 2)} - \frac{x^{2} - 4 x - 6}{(x + 6) (x - 2)} \\ The denominators are the same . Combine the numerators being careful to note the negative sign . \\ \frac{3 x^{2} + 4 x + 5 + 2 x^{2} + x + 6 - (x^{2} - 4 x + 6)}{(x + 6) (x - 2)} \\ \frac{3 x^{2} + 4 x + 5 + 2 x^{2} + x + 6 - x^{2} + 4 x + 6}{(x + 6) (x - 2)} \\ \frac{4 x^{2} + 9 x + 17}{(x + 6) (x - 2)} \end{array}$

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Practice set a

Add or Subtract the following rational expressions.

$\frac{4}{9} + \frac{2}{9}$

$\frac{2}{3}$

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$\frac{3}{b} + \frac{2}{b}$

$\frac{5}{b}$

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$\frac{5 x}{2 y^{2}} - \frac{3 x}{2 y^{2}}$

$\frac{x}{y^{2}}$

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$\frac{x + y}{x - y} + \frac{2 x + 3 y}{x - y}$

$\frac{3 x + 4 y}{x - y}$

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$\frac{4 x^{2} - x + 4}{3 x + 10} - \frac{x^{2} + 2 x + 5}{3 x + 10}$

$\frac{3 x^{2} - 3 x - 1}{3 x + 10}$

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$\frac{x (x + 1)}{x (2 x + 3)} + \frac{3 x^{2} - x + 7}{2 x^{2} + 3 x}$

$\frac{4 x^{2} + 7}{x (2 x + 3)}$

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$\frac{4 x + 3}{x^{2} - x - 6} - \frac{8 x - 4}{(x + 2) (x - 3)}$

$\frac{- 4 x + 7}{(x + 2) (x - 3)}$

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$\frac{5 a^{2} + a - 4}{2 a (a - 6)} + \frac{2 a^{2} + 3 a + 4}{2 a^{2} - 12 a} + \frac{a^{2} + 2}{2 a^{2} - 12 a}$

$\frac{4 a^{2} + 2 a + 1}{a (a - 6)}$

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$\frac{8 x^{2} + x - 1}{x^{2} - 6 x + 8} + \frac{2 x^{2} + 3 x}{x^{2} - 6 x + 8} - \frac{5 x^{2} + 3 x - 4}{(x - 4) (x - 2)}$

$\frac{5 x^{2} + x + 3}{(x - 4) (x - 2)}$

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Fractions with different denominators

Sample set b

Add or Subtract the following rational expressions.

$\begin{array}{l} \frac{4 a}{3 y} + \frac{2 a}{9 y^{2}} . & The denominators are n o t the same . Find the LCD . By inspection, the LCD is 9 y^{2} . \\ \frac{}{9 y^{2}} + \frac{2 a}{9 y^{2}} & \begin{array}{l} The denominator of the first rational expression has been multiplied by 3 y, \\ so the numerator must be multiplied by 3 y . \\ 4 a \cdot 3 y = 12 a y \end{array} \\ \frac{12 a y}{9 y^{2}} + \frac{2 a}{9 y^{2}} & The denominators are now the same . Add the numerators . \\ \frac{12 a y + 2 a}{9 y^{2}} \end{array}$

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$\begin{array}{l} \frac{3 b}{b + 2} + \frac{5 b}{b - 3} . & \begin{array}{l} The denominators are n o t the same . The LCD is (b + 2) (b - 3) . \end{array} \\ \frac{}{(b + 2) (b - 3)} + \frac{}{(b + 2) (b - 3)} & \begin{array}{l} The denominator of the first rational expression has been multiplied by b - 3, \\ so the numerator must be multiplied by b - 3. 3 b (b - 3) \end{array} \\ \frac{3 b (b - 3)}{(b + 2) (b - 3)} + \frac{}{(b + 2) (b - 3)} & \begin{array}{l} The denominator of the second rational expression has been multiplied by b + 2, \\ so the numerator must be multiplied by b + 2. 5 b (b + 2) \end{array} \\ \frac{3 b (b - 3)}{(b + 2) (b - 3)} + \frac{5 b (b + 2)}{(b + 2) (b - 3)} & \begin{array}{l} The denominators are now the same . Add the numerators . \end{array} \\ \frac{3 b (b - 3) + 5 b (b + 2)}{(b - 3) (b + 2)} & = & \frac{3 b^{2} - 9 b + 5 b^{2} + 10 b}{(b - 3) (b + 2)} \\ = & \frac{8 b^{2} + b}{(b - 3) (b - 2)} \end{array}$

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$\begin{array}{l} \frac{x + 3}{x - 1} + \frac{x - 2}{4 x + 4} . & \begin{array}{l} The denominators are n o t the same . \\ Find the LCD . \end{array} \\ \frac{x + 3}{x - 1} + \frac{x - 2}{4 (x + 1)} & The LCD is (x + 1) (x - 1) \\ \frac{}{4 (x + 1) (x - 1)} + \frac{}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominator of the first rational expression has been multiplied by 4 (x + 1) so \\ the numerator must be multiplied by 4 (x + 1) . 4 (x + 3) (x + 1) \end{array} \\ \frac{4 (x + 3) (x + 1)}{4 (x + 1) (x - 1)} + \frac{}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominator of the second rational expression has been multiplied by x - 1 \\ so the numerator must be multiplied by x - 1. (x - 1) (x - 2) \end{array} \\ \frac{4 (x + 3) (x + 1)}{4 (x + 1) (x - 1)} + \frac{(x - 1) (x - 2)}{4 (x + 1) (x - 1)} & \begin{array}{l} The denominators are now the same. \\ Add the numerators. \end{array} \\ \frac{4 (x + 3) (x + 1) + (x - 1) (x - 2)}{4 (x + 1) (x - 1)} \\ \frac{4 (x^{2} + 4 x + 3) + x^{2} - 3 x + 2}{4 (x + 1) (x - 1)} \\ \frac{4 x^{2} + 16 x + 12 + x^{2} - 3 x + 2}{4 (x + 1) (x - 1)} & = & \frac{5 x^{2} + 13 x + 14}{4 (x + 1) (x - 1)} \end{array}$

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$\begin{array}{l} \frac{x + 5}{x^{2} - 7 x + 12} + \frac{3 x - 1}{x^{2} - 2 x - 3} & Determine the LCD . \\ \frac{x + 5}{(x - 4) (x - 3)} + \frac{3 x - 1}{(x - 3) (x + 1)} & The LCD is (x - 4) (x - 3) (x + 1) . \\ \frac{}{(x - 4) (x - 3) (x + 1)} + \frac{}{(x - 4) (x - 3) (x + 1)} & The first numerator must be multiplied by x + 1 and the second by x - 4. \\ \frac{(x + 5) (x + 1)}{(x - 4) (x - 3) (x + 1)} + \frac{(3 x - 1) (x - 4)}{(x - 4) (x - 3) (x + 1)} & The denominators are now the same . Add the numerators . \\ \frac{(x + 5) (x + 1) + (3 x - 1) (x - 4)}{(x - 4) (x - 3) (x + 1)} \\ \frac{x^{2} + 6 x + 5 + 3 x^{2} - 13 x + 4}{(x - 4) (x - 3) (x + 1)} \\ \frac{4 x^{2} - 7 x + 9}{(x - 4) (x - 3) (x + 1)} \end{array}$

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$\begin{array}{l} \frac{a + 4}{a^{2} + 5 a + 6} - \frac{a - 4}{a^{2} - 5 a - 24} & Determine the LCD. \\ \frac{a + 4}{(a + 3) (a + 2)} - \frac{a - 4}{(a + 3) (a - 8)} & The LCD is (a + 3) (a + 2) (a - 8) . \\ \frac{}{(a + 3) (a + 2) (a - 8)} - \frac{}{(a + 3) (a + 2) (a - 8)} & The first numerator must be multiplied by a - 8 and the second by a + 2. \\ \frac{(a + 4) (a - 8)}{(a + 3) (a + 2) (a - 8)} - \frac{(a - 4) (a + 2)}{(a + 3) (a + 2) (a - 8)} & The denominators are now the same . Subtract the numerators . \\ \frac{(a + 4) (a - 8) - (a - 4) (a + 2)}{(a + 3) (a + 2) (a - 8)} \\ \frac{a^{2} - 4 a - 32 - (a^{2} - 2 a - 8)}{(a + 3) (a + 2) (a - 8)} \\ \frac{a^{2} - 4 a - 32 - a^{2} + 2 a + 8}{(a + 3) (a + 2) (a - 8)} \\ \frac{- 2 a - 24}{(a + 3) (a + 2) (a - 8)} & Factor - 2 from the numerator . \\ \frac{- 2 (a + 12)}{(a + 3) (a + 2) (a - 8)} \end{array}$

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$\begin{array}{l} \frac{3 x}{7 - x} + \frac{5 x}{x - 7} . & \begin{array}{l} The denominators are n e a r l y the same. They differ only in sign. \\ Our technique is to factor - 1 from one of them. \end{array} \\ \frac{3 x}{7 - x} = \frac{3 x}{- (x - 7)} & = & \frac{- 3 x}{x - 7} & Factor - 1 from the first term . \\ \frac{3 x}{7 - x} + \frac{5 x}{x - 7} & = & \frac{- 3 x}{x - 7} + \frac{5 x}{x - 7} \\ = & \frac{- 3 x + 5 x}{x - 7} \\ = & \frac{2 x}{x - 7} \end{array}$

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Practice set b

Add or Subtract the following rational expressions.

$\frac{3 x}{4 a^{2}} + \frac{5 x}{12 a^{3}}$

$\frac{9 a x + 5 x}{12 a^{3}}$

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$\frac{5 b}{b + 1} + \frac{3 b}{b - 2}$

$\frac{8 b^{2} - 7 b}{(b + 1) (b - 2)}$

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$\frac{a - 7}{a + 2} + \frac{a - 2}{a + 3}$

$\frac{2 a^{2} - 4 a - 25}{(a + 2) (a + 3)}$

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$\frac{4 x + 1}{x + 3} - \frac{x + 5}{x - 3}$

$\frac{3 x^{2} - 19 x - 18}{(x + 3) (x - 3)}$

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$\frac{2 y - 3}{y} + \frac{3 y + 1}{y + 4}$

$\frac{5 y^{2} + 6 y - 12}{y (y + 4)}$

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$\frac{a - 7}{a^{2} - 3 a + 2} + \frac{a + 2}{a^{2} - 6 a + 8}$

$\frac{2 a^{2} - 10 a + 26}{(a - 2) (a - 1) (a - 4)}$

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$\frac{6}{b^{2} + 6 b + 9} - \frac{2}{b^{2} + 4 b + 4}$

$\frac{4 b^{2} + 12 b + 6}{{(b + 3)}^{2} {(b + 2)}^{2}}$

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$\frac{x}{x + 4} - \frac{x - 2}{3 x - 3}$

$\frac{2 x^{2} - 5 x + 8}{3 (x + 4) (x - 1)}$

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$\frac{5 x}{4 - x} + \frac{7 x}{x - 4}$

$\frac{2 x}{x - 4}$

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Sample set c

Combine the following rational expressions.

$\begin{array}{l} 3 + \frac{7}{x - 1} . & Rewrite the expression. \\ \frac{3}{1} + \frac{7}{x - 1} & The LCD is x - 1. \\ \frac{3 (x - 1)}{x - 1} + \frac{7}{x - 1} & = & \frac{3 x - 3}{x - 1} + \frac{7}{x - 1} & = & \frac{3 x - 3 + 7}{x - 1} \\ = & \frac{3 x + 4}{x - 1} \end{array}$

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$\begin{array}{l} 3 y + 4 - \frac{y^{2} - y + 3}{y - 6} . & Rewrite the expression. \\ \frac{3 y + 4}{1} - \frac{y^{2} - y + 3}{y - 6} & The LCD is y - 6. \\ \frac{(3 y + 4) (y - 6)}{y - 6} - \frac{y^{2} - y + 3}{y - 6} & = & \frac{(3 y + 4) (y - 6) - (y^{2} - y + 3)}{y - 6} \\ = & \frac{3 y^{2} - 14 y - 24 - y^{2} + y - 3}{y - 6} \\ = & \frac{2 y^{2} - 13 y - 27}{y - 6} \end{array}$

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Practice set c

Simplify $8 + \frac{3}{x - 6} .$

$\frac{8 x - 45}{x - 6}$

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Simplify $2 a - 5 - \frac{a^{2} + 2 a - 1}{a + 3} .$

$\frac{a^{2} - a - 14}{a + 3}$

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Exercises

For the following problems, add or subtract the rational expressions.

$\frac{3}{8} + \frac{1}{8}$

$\frac{1}{2}$

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$\frac{1}{9} + \frac{4}{9}$

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$\frac{7}{10} - \frac{2}{5}$

$\frac{3}{10}$

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$\frac{3}{4} - \frac{5}{12}$

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$\frac{3}{4 x} + \frac{5}{4 x}$

$\frac{2}{x}$

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$\frac{2}{7 y} + \frac{3}{7 y}$

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$\frac{6 y}{5 x} + \frac{8 y}{5 x}$

$\frac{14 y}{5 x}$

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$\frac{9 a}{7 b} + \frac{3 a}{7 b}$

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$\frac{15 n}{2 m} - \frac{6 n}{2 m}$

$\frac{9 n}{2 m}$

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$\frac{8 p}{11 q} - \frac{3 p}{11 q}$

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$\frac{y + 4}{y - 6} + \frac{y + 8}{y - 6}$

$\frac{2 y + 12}{y - 6}$

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$\frac{y - 1}{y + 4} + \frac{y + 7}{y + 4}$

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$\frac{a + 6}{a - 1} + \frac{3 a + 5}{a - 1}$

$\frac{4 a + 11}{a - 1}$

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$\frac{5 a + 1}{a + 7} + \frac{2 a - 6}{a + 7}$

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$\frac{x + 1}{5 x} + \frac{x + 3}{5 x}$

$\frac{2 x + 4}{5 x}$

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$\frac{a - 6}{a + 2} + \frac{a - 2}{a + 2}$

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$\frac{b + 1}{b - 3} + \frac{b + 2}{b - 3}$

$\frac{2 b + 3}{b - 3}$

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$\frac{a + 2}{a - 5} - \frac{a + 3}{a - 5}$

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$\frac{b + 7}{b - 6} - \frac{b - 1}{b - 6}$

$\frac{8}{b - 6}$

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$\frac{2 b + 3}{b + 1} - \frac{b - 4}{b + 1}$

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$\frac{3 y + 4}{y + 8} - \frac{2 y - 5}{y + 8}$

$\frac{y + 9}{y + 8}$

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$\frac{2 a - 7}{a - 9} + \frac{3 a + 5}{a - 9}$

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$\frac{8 x - 1}{x + 2} - \frac{15 x + 7}{x + 2}$

$\frac{- 7 x - 8}{x + 2}$

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$\frac{7}{2 x^{2}} + \frac{1}{6 x^{3}}$

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$\frac{2}{3 x} + \frac{4}{6 x^{2}}$

$\frac{2 (x + 1)}{3 x^{2}}$

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$\frac{5}{6 y^{3}} - \frac{2}{18 y^{5}}$

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$\frac{2}{5 a^{2}} - \frac{1}{10 a^{3}}$

$\frac{4 a - 1}{10 a^{3}}$

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$\frac{3}{x + 1} + \frac{5}{x - 2}$

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$\frac{4}{x - 6} + \frac{1}{x - 1}$

$\frac{5 (x - 2)}{(x - 6) (x - 1)}$

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$\frac{2 a}{a + 1} - \frac{3 a}{a + 4}$

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$\frac{6 y}{y + 4} + \frac{2 y}{y + 3}$

$\frac{2 y (4 y + 13)}{(y + 4) (y + 3)}$

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$\frac{x - 1}{x - 3} + \frac{x + 4}{x - 4}$

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$\frac{x + 2}{x - 5} + \frac{x - 1}{x + 2}$

$\frac{2 x^{2} - 2 x + 9}{(x - 5) (x + 2)}$

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$\frac{a + 3}{a - 3} - \frac{a + 2}{a - 2}$

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$\frac{y + 1}{y - 1} - \frac{y + 4}{y - 4}$

$\frac{- 6 y}{(y - 1) (y - 4)}$

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$\frac{x - 1}{(x + 2) (x - 3)} + \frac{x + 4}{x - 3}$

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$\frac{y + 2}{(y + 1) (y + 6)} + \frac{y - 2}{y + 6}$

$\frac{y^{2}}{(y + 1) (y + 6)}$

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$\frac{2 a + 1}{(a + 3) (a - 3)} - \frac{a + 2}{a + 3}$

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$\frac{3 a + 5}{(a + 4) (a - 1)} - \frac{2 a - 1}{a - 1}$

$\frac{- 2 a^{2} - 4 a + 9}{(a + 4) (a - 1)}$

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$\frac{2 x}{x^{2} - 3 x + 2} + \frac{3}{x - 2}$

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$\frac{4 a}{a^{2} - 2 a - 3} + \frac{3}{a + 1}$

$\frac{7 a - 9}{(a + 1) (a - 3)}$

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$\frac{3 y}{y^{2} - 7 y + 12} - \frac{y^{2}}{y - 3}$

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$\frac{x - 1}{x^{2} + 6 x + 8} + \frac{x + 3}{x^{2} + 2 x - 8}$

$\frac{2 (x^{2} + x + 4)}{(x + 2) (x - 2) (x + 4)}$

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$\frac{a - 4}{a^{2} + 2 a - 3} + \frac{a + 2}{a^{2} + 3 a - 4}$

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$\frac{b - 3}{b^{2} + 9 b + 20} + \frac{b + 4}{b^{2} + b - 12}$

$\frac{2 b^{2} + 3 b + 29}{(b - 3) (b + 4) (b + 5)}$

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$\frac{y - 1}{y^{2} + 4 y - 12} - \frac{y + 3}{y^{2} + 6 y - 16}$

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$\frac{x + 3}{x^{2} + 9 x + 14} - \frac{x - 5}{x^{2} - 4}$

$\frac{- x + 29}{(x - 2) (x + 2) (x + 7)}$

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$\frac{x - 1}{x^{2} - 4 x + 3} + \frac{x + 3}{x^{2} - 5 x + 6} + \frac{2 x}{x^{2} - 3 x + 2}$

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$\frac{4 x}{x^{2} + 6 x + 8} + \frac{3}{x^{2} + x - 6} + \frac{x - 1}{x^{2} + x - 12}$

$\frac{5 x^{4} - 3 x^{3} - 34 x^{2} + 34 x - 60}{(x - 2) (x + 2) (x - 3) (x + 3) (x + 4)}$

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$\frac{y + 2}{y^{2} - 1} + \frac{y - 3}{y^{2} - 3 y - 4} - \frac{y + 3}{y^{2} - 5 y + 4}$

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$\frac{a - 2}{a^{2} - 9 a + 18} + \frac{a - 2}{a^{2} - 4 a - 12} - \frac{a - 2}{a^{2} - a - 6}$

$\frac{(a + 5) (a - 2)}{(a + 2) (a - 3) (a - 6)}$

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$\frac{y - 2}{y^{2} + 6 y} + \frac{y + 4}{y^{2} + 5 y - 6}$

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$\frac{a + 1}{a^{3} + 3 a^{2}} - \frac{a + 6}{a^{2} - a}$

$\frac{- a^{3} - 8 a^{2} - 18 a - 1}{a^{2} (a + 3) (a - 1)}$

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$\frac{4}{3 b^{2} - 12 b} - \frac{2}{6 b^{2} - 6 b}$

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$\frac{3}{2 x^{5} - 4 x^{4}} + \frac{- 2}{8 x^{3} + 24 x^{2}}$

$\frac{- x^{3} + 2 x^{2} + 6 x + 18}{4 x^{4} (x - 2) (x + 3)}$

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$\frac{x + 2}{12 x^{3}} + \frac{x + 1}{4 x^{2} + 8 x - 12} - \frac{x + 3}{16 x^{2} - 32 x + 16}$

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$\frac{2 x}{x^{2} - 9} - \frac{x + 1}{4 x^{2} - 12 x} - \frac{x - 4}{8 x^{3}}$

$\frac{14 x^{4} - 9 x^{3} - 2 x^{2} + 9 x - 36}{8 x^{3} (x + 3) (x - 3)}$

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$4 + \frac{3}{x + 2}$

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$8 + \frac{2}{x + 6}$

$\frac{8 x + 50}{x + 6}$

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$1 + \frac{4}{x - 7}$

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$3 + \frac{5}{x - 6}$

$\frac{3 x - 13}{x - 6}$

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$- 2 + \frac{4 x}{x + 5}$

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$- 1 + \frac{3 a}{a - 1}$

$\frac{2 a + 1}{a - 1}$

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$6 - \frac{4 y}{y + 2}$

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$2 x + \frac{x^{2} - 4}{x + 1}$

$\frac{3 x^{2} + 2 x - 4}{x + 1}$

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$- 3 y + \frac{4 y^{2} + 2 y - 5}{y + 3}$

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$x + 2 + \frac{x^{2} + 4}{x - 1}$

$\frac{2 x^{2} + x + 2}{x - 1}$

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$b + 6 + \frac{2 b + 5}{b - 2}$

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$\frac{3 x - 1}{x - 4} - 8$

$\frac{- 5 x + 31}{x - 4}$

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$\frac{4 y + 5}{y + 1} - 9$

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$\frac{2 y^{2} + 11 y - 1}{y + 4} - 3 y$

$\frac{- (y^{2} + y + 1)}{y + 4}$

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$\frac{5 y^{2} - 2 y + 1}{y^{2} + y - 6} - 2$

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$\frac{4 a^{3} + 2 a^{2} + a - 1}{a^{2} + 11 a + 28} + 3 a$

$\frac{7 a^{3} + 35 a^{2} + 85 a - 1}{(a + 7) (a + 4)}$

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$\frac{2 x}{1 - x} + \frac{6 x}{x - 1}$

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$\frac{5 m}{6 - m} + \frac{3 m}{m - 6}$

$\frac{- 2 m}{m - 6}$

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$\frac{- a + 7}{8 - 3 a} + \frac{2 a + 1}{3 a - 8}$

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$\frac{- 2 y + 4}{4 - 5 y} - \frac{9}{5 y - 4}$

$\frac{2 y - 13}{5 y - 4}$

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$\frac{m - 1}{1 - m} - \frac{2}{m - 1}$

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Exercises for review

( [link] ) Simplify ${(x^{3} y^{2} z^{5})}^{6} {(x^{2} y z)}^{2} .$

$x^{22} y^{14} z^{32}$

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( [link] ) Write $6 a^{- 3} b^{4} c^{- 2} a^{- 1} b^{- 5} c^{3}$ so that only positive exponents appear.

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( [link] ) Construct the graph of $y = - 2 x + 4.$
An xy coordinate plane with gridlines, labeled negative five to five on both axes.

A graph of a line passing through three points with coordinates zero, four; one, two; and two, zero.

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( [link] ) Find the product: $\frac{x^{2} - 3 x - 4}{x^{2} + 6 x + 5} \cdot \frac{x^{2} + 5 x + 6}{x^{2} - 2 x - 8} .$

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( [link] ) Replace N with the proper quantity: $\frac{x + 3}{x - 5} = \frac{N}{x^{2} - 7 x + 10} .$

$(x + 3) (x - 2)$

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hey

Sammol

God

hello

Jahara

Good morning

Jorge

abubakar

Nmesoma

Mahesh

Tom

Why is unemployment rate never zero at full employment?

Priyanka Reply

bcoz of existence of frictional unemployment in our economy.

Umashankar

what is flexible exchang rate?

poudel

due to existence of the pple with disabilities

Abdulraufu

the demand of a good rises, causing the demand for another good to fall

Rushawn Reply

is it possible to leave every good at the same level

Joseph

I don't think so. because check it, if the demand for chicken increases, people will no longer consume fish like they used to causing a fall in the demand for fish

Anuolu

is not really possible to let the value of a goods to be same at the same time.....

Salome

Suppose the inflation rate is 6%, does it mean that all the goods you purchase will cost 6% more than previous year? Provide with reasoning.

Geetha Reply

Not necessarily. To measure the inflation rate economists normally use an averaged price index of a basket of certain goods. So if you purchase goods included in the basket, you will notice that you pay 6% more, otherwise not necessarily.

Waeth

discus major problems of macroeconomics

Alii Reply

what is the problem of macroeconomics

Yoal

Economic growth Stable prices and low unemployment

Ephraim

explain inflationcause and itis degre

Miresa Reply

what is inflation

Getu

increase in general price levels

WEETO

Good day How do I calculate this question: C= 100+5yd G= 2000 T= 2000 I(planned)=200. Suppose the actual output is 3000. What is the level of planned expenditures at this level of output?

Chisomo Reply

how to calculate actual output?

Chisomo

how to calculate the equilibrium income

Beshir

Criteria for determining money supply

Thapase Reply

who we can define macroeconomics in one line

Muhammad

Aggregate demand

Mohammed

C=k100 +9y and i=k50.calculate the equilibrium level of output

Mercy Reply

Isiaka

Geli

Man

👋

Bahunda

hy how are you?

Man

Amisha

how are you guys

Sekou

f9 guys

Amisha

how are you guys

Sekou

ys am also fine

Amisha

fine and you guys

Geli

from Nepal

Amisha

nawalparasi district from belatari

Amisha

nd u

Amisha

I am Camara from Guinea west Africa... happy to meet you guys here

Sekou

ma management ho

Amisha

ahile becheclor ho

Amisha

hjr ktm bta ho ani k kaam grnu hunxa tw

Amisha

belatari

Amisha

1st year ho

Amisha

nd u

Amisha

ahh

Amisha

kaha biratnagar

Amisha

kina k vo

Amisha

money as unit of account means what?

Kalombe

A unit of account is something that can be used to value goods and services and make calculations

Jim

all of you please speak in English I can't understand you're language

Muhammad

I want to know how can we define macroeconomics in one line

Muhammad

it must be .9 or 0.9 no Mpc is greater than 1 Y=100+.9Y+50 Y-.9Y=150 0.1Y/0.1=150/0.1 Y=1500

Kalombe

Mercy is it clear?😋

Kalombe

hi can someone help me on this question If a negative shocks shifts the IS curve to the left, what type of policy do you suggest so as to stabilize the level of output? discuss your answer using appropriate graph.

Galge Reply

if interest rate is increased this will will reduce the level of income shifting the curve to the left ◀️

Kalombe

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Source: OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3

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