Answer:
here's the solution : -
=》2x - 3y = -29
=》x + 6y = 23
=》x = 23 - 6y
now, plugging the value of x in equation 1, we get : -
=》2x - 3y = -29
=》2 (23 - 6y) - 3y = - 29
=》46 - 12y - 3y = -29
=》-15y = -29 - 46
=》y = -75 ÷ -15
=》y = 5
now, plugging the value of y as 5 in equation 2
=》x + 6y = 23
=》x + (6 × 5.2) = 23
=》x = 23 - 30
=》x = - 7
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Answer:
x^2+8x+<u>1</u><u>6</u><u>=</u><u>(</u><u>x-4</u><u>)</u><u>^</u><u>2</u>
<em><u>EXPLANATION</u></em><em><u>:</u></em>
<u>(</u><u>a</u><u>+</u><u>b</u><u>)</u><u>^</u><u>2</u><u>=</u><u>a2</u><u>+</u><u>2</u><u>.</u><u>a</u><u>.</u><u>b</u><u>+</u><u>b2</u>
<u>we</u><u> </u><u>have</u><u> </u><u>to</u><u> </u><u>break</u><u> </u><u>the</u><u> </u><u>middle</u><u> </u><u>term</u><u> </u><u>i</u><u>n</u><u> </u><u>2</u><u>a</u><u>b</u><u> </u><u>here</u><u> </u><u>a</u><u> </u><u>is</u><u> </u><u>x</u><u> </u><u>then</u><u> </u><u>2</u><u>x</u><u>b</u><u>=</u><u>8</u><u>x</u><u>,</u><u> </u><u>=</u><u>></u><u> </u><u>b</u><u>=</u><u>4</u><u>,</u><u> </u><u>but</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>and</u><u> </u><u>b</u><u> </u><u>to</u><u> </u><u>get</u><u> </u><u>the</u><u> </u><u>req</u><u>uired</u><u> </u><u>equation</u><u>!</u>
<em>Given, 9−7x=5−3x</em>
<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.</em>
<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x</em>
<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x⇒4=4x</em>
<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x⇒4=4x⇒x= </em><em>4</em><em>/</em><em>4</em><em> </em><em> =1</em>
<em> =1Hence, required solution is x=1.</em>
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