{"id":"cb_question_bank:ca041044-8afe-4361-91b7-3cb54b318d2f","source_id":"cb_question_bank","source_item_id":"ca041044-8afe-4361-91b7-3cb54b318d2f","external_id":"f70845bb-9302-4075-b020-3ece6ab22306","ibn":null,"question_id":"bef4b1c6","program":"SAT","module":"math","domain_code":"P","domain":"Advanced Math","skill_code":"P.B.","skill":"Nonlinear equations in one variable and systems of equations in two variables","difficulty":"M","score_band":4,"answer_type":"spr","answer_source":"key","layout_warning":null,"form_id":null,"module_number":null,"position":null,"stem":"StartFraction 55 Over x plus 6 EndFraction equals x What is the positive solution to the given equation?","stimulus":"","stimulus_html":null,"stem_html":"<p style=\"text-align: center;\"><math alttext=\"StartFraction 55 Over x plus 6 EndFraction equals x\"><mrow>\n\t<mrow>\n\t\t<mfrac>\n\t\t\t<mn>55</mn>\n\t\t\t<mrow>\n\t\t\t\t<mi>x</mi>\n\t\t\t\t<mo>+</mo>\n\t\t\t\t<mn>6</mn>\n\t\t\t</mrow>\n\t\t</mfrac>\n\t</mrow>\n\t<mo>=</mo>\n\t<mi>x</mi>\n</mrow>\n</math></p>\n<p style=\"text-align: left;\">What is the positive solution to the given equation?</p>","rationale_html":"<p style=\"text-align: left;\">The correct answer is <math alttext=\"5\"><mn>5</mn>\n</math>. Multiplying both sides of the given equation by <math alttext=\"x plus 6\"><mrow>\n\t<mi>x</mi>\n\t<mo>+</mo>\n\t<mn>6</mn>\n</mrow>\n</math> results in <math alttext=\"55 equals x left parenthesis x plus 6 right parenthesis\"><mn>55</mn><mo>=</mo><mi>x</mi><mfenced><mrow><mi>x</mi><mo>+</mo><mn>6</mn></mrow></mfenced></math>. Applying the distributive property of multiplication to the right-hand side of this equation results in <math alttext=\"55 equals x squared plus 6 x\"><mrow>\n\t<mn>55</mn>\n\t<mo>=</mo>\n\t<mrow>\n\t\t<msup>\n\t\t\t<mi>x</mi>\n\t\t\t<mn>2</mn>\n\t\t</msup>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>6</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t</mrow>\n</mrow>\n</math>. Subtracting <math alttext=\"55\"><mn>55</mn>\n</math> from both sides of this equation results in <math alttext=\"0 equals x squared plus 6 x minus 55\"><mrow>\n\t<mn>0</mn>\n\t<mo>=</mo>\n\t<mrow>\n\t\t<msup>\n\t\t\t<mi>x</mi>\n\t\t\t<mn>2</mn>\n\t\t</msup>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>6</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>-</mo>\n\t\t<mn>55</mn>\n\t</mrow>\n</mrow>\n</math>. The right-hand side of this equation can be rewritten by factoring. The two values that multiply to <math alttext=\"negative 55\"><mo>-</mo><mn>55</mn>\n</math> and add to <math alttext=\"6\"><mn>6</mn>\n</math> are <math alttext=\"11\"><mn>11</mn>\n</math> and <math alttext=\"negative 5\"><mo>-</mo><mn>5</mn>\n</math>. It follows that the equation <math alttext=\"0 equals x squared plus 6 x minus 55\"><mrow>\n\t<mn>0</mn>\n\t<mo>=</mo>\n\t<mrow>\n\t\t<msup>\n\t\t\t<mi>x</mi>\n\t\t\t<mn>2</mn>\n\t\t</msup>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>6</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>-</mo>\n\t\t<mn>55</mn>\n\t</mrow>\n</mrow>\n</math> can be rewritten as <math alttext=\"0 equals left parenthesis x plus 11 right parenthesis left parenthesis x minus 5 right parenthesis\"><mn>0</mn><mo>=</mo><mfenced><mrow><mi>x</mi><mo>+</mo><mn>11</mn></mrow></mfenced><mfenced><mrow><mi>x</mi><mo>-</mo><mn>5</mn></mrow></mfenced></math>. Setting each factor equal to <math alttext=\"0\"><mn>0</mn>\n</math> yields two equations: <math alttext=\"x plus 11 equals 0\"><mrow>\n\t<mrow>\n\t\t<mi>x</mi>\n\t\t<mo>+</mo>\n\t\t<mn>11</mn>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>0</mn>\n</mrow>\n</math> and <math alttext=\"x minus 5 equals 0\"><mrow>\n\t<mrow>\n\t\t<mi>x</mi>\n\t\t<mo>-</mo>\n\t\t<mn>5</mn>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>0</mn>\n</mrow>\n</math>. Subtracting <math alttext=\"11\"><mn>11</mn>\n</math> from both sides of the equation <math alttext=\"x plus 11 equals 0\"><mrow>\n\t<mrow>\n\t\t<mi>x</mi>\n\t\t<mo>+</mo>\n\t\t<mn>11</mn>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>0</mn>\n</mrow>\n</math> results in <math alttext=\"x equals negative 11\"><mrow>\n\t<mi>x</mi>\n\t<mo>=</mo>\n\t<mo>-</mo><mn>11</mn>\n</mrow>\n</math>. Adding <math alttext=\"5\"><mn>5</mn>\n</math> to both sides of the equation <math alttext=\"x minus 5 equals 0\"><mrow>\n\t<mrow>\n\t\t<mi>x</mi>\n\t\t<mo>-</mo>\n\t\t<mn>5</mn>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>0</mn>\n</mrow>\n</math> results in <math alttext=\"x equals 5\"><mrow>\n\t<mi>x</mi>\n\t<mo>=</mo>\n\t<mn>5</mn>\n</mrow>\n</math>. Therefore, the positive solution to the given equation is <math alttext=\"5\"><mn>5</mn>\n</math>.</p>","correct_answer":["5"],"has_media":false,"has_mathml":true,"has_table":false,"source_created_at":"2023-08-02T20:25:59.823000+00:00","source_updated_at":"2023-08-02T20:25:59.823000+00:00","options":[],"attribution":{"source_id":"cb_question_bank","source_name":"College Board digital SAT question bank (community dump)","rights_holder":"College Board","canonical_url":"https://satsuitequestionbank.collegeboard.org/","retrieved_from":"https://raw.githubusercontent.com/mdn522/sat-question-bank/main/data/cb-digital-questions.json","attribution":"SAT practice questions © College Board, from the official SAT Suite Question Bank. Retrieved via the community dump at github.com/mdn522/sat-question-bank.","disclaimer":"SAT® and PSAT/NMSQT® are trademarks registered by the College Board, which is not affiliated with, does not sponsor, and does not endorse this project.","license":"College Board copyright; source repository carries no licence. Local use only.","redistributable":false,"item_count":2017}}