Envision Algebra 2 2-7 Additional Practice: Linear-Quadratic Systems Determine the number of solutions for the system of equations: Y = -x^2 + 3x + 2 y = -x^2 + 2x + 18 y = x + 3x – 5 (y = 3x + 2 y = 5x – 10 Y = -x^2 – Zx + 1) Use substitution to solve the system of equations: Sy = x^2 + Sx – 2 Ky = -x^2 + 12 ly = 3x – 2 (y = 2x – 8) y = x – 2x – 3 (y = 2 – 3) Solve each system of inequalities using shading: Sy > 3x^2 + 3x – 5 fy > 4x^2 + 8x + 8 ly < -3x – 5 4x + 8 > y In business, the break-even point is the point (x) at which the graphs of the revenue and cost functions intersect. This is the point where the revenue and cost are equal.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Answers<\/h3>\n\n\n\n
Problem<\/th>
Equations or Inequalities<\/th>
Solutions \/ Region<\/th>
# of real solutions*<\/th><\/tr><\/thead>
A<\/strong><\/td>
y=\u2212x2+3x+2y=-x^{2}+3x+2 and y=\u2212x2+2x+18y=-x^{2}+2x+18<\/td>
y=4x\u22125y=4x-5 and y=\u2212x2\u22122x+1y=-x^{2}-2x+1<\/td>
x=\u22123\u00b115\u2005\u200a\u2005\u200a(y=4x\u22125)x=-3\\pm\\sqrt{15}\\;\\;\\bigl(y=4x-5\\bigr) \u2248 (0.873,\u2009\u22121.51)(0.873,\\,-1.51) and (\u2009\u22126.873,\u2009\u221232.49)(\\,-6.873,\\,-32.49)<\/td>
2<\/strong><\/td><\/tr>
C<\/strong><\/td>
y=3x+2y=3x+2 and y=5x\u221210y=5x-10<\/td>
(6,\u200920)(6,\\,20)<\/td>
1<\/strong><\/td><\/tr>
D<\/strong><\/td>
y=x2+5x\u22122y=x^{2}+5x-2 and y=\u2212x2+12y=-x^{2}+12<\/td>
Shade the set above<\/strong> both parabolas and<\/em> below<\/strong> both lines. Region is closed on the left and right by the lines, and bounded beneath by the higher of the two parabolas.<\/td>
<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
*\u201c# of real solutions\u201d refers to the number of intersection points of each system.<\/p>\n\n\n\n
\n\n\n\n
explanation<\/h3>\n\n\n\n
A linear\u2013quadratic system is solved by substitution<\/strong> because both equations already give yy explicitly. Setting the right-hand sides equal makes the common yy disappear and leaves an equation in one variable.<\/p>\n\n\n\n
Problem A<\/strong> shows why the count of solutions can sometimes be only one. Both equations share the same \u2212x2-x^{2} term, so it cancels entirely, leaving the linear<\/em> equation 3x+2=2x+183x+2=2x+18. A linear equation has a single root, so exactly one ordered pair satisfies both formulas. This single point is their only intersection.<\/p>\n\n\n\n
In Problem B<\/strong> the linear graph y=4x\u22125y=4x-5 meets the downward-opening parabola y=\u2212x2\u22122x+1y=-x^{2}-2x+1. Equating them produces the quadratic x2+6x\u22126=0x^{2}+6x-6=0. Because its discriminant b2\u22124ac=60b^{2}-4ac=60 is positive, two distinct real roots emerge. Each root gives a different xx; plugging back into either original equation supplies the two corresponding yy-values. Therefore the graphs cross twice\u2014two solutions. Whenever the discriminant is zero the graphs would just kiss once (a tangent), and when it is negative they never meet, yielding zero real solutions.<\/p>\n\n\n\n
Problems C\u2013F<\/strong> repeat the same maneuver with different pairings (linear\u2013linear or quadratic\u2013quadratic). Every time you equate the right-hand sides, simplify, and examine the resulting polynomial\u2019s discriminant you know instantly how many intersection points to expect. After that, substituting each root back retrieves the complete ordered pairs.<\/p>\n\n\n\n
For the inequality system<\/strong> we treat each equation as a boundary curve. Arrows on a sketch help: opening upward parabolas form the bottom of the feasible set because we need yy greater<\/em>. The two straight lines, both slanted upward, cap the region because we simultaneously need yy less<\/em>. The physically shaded area on a graphing page is the overlap of those four half-planes: above both curves, below both lines. Identifying such an overlap is exactly what businesses do when they graph cost and revenue: the single point where R(x)=C(x)R(x)=C(x) is the break-even<\/em>\u2014the same algebraic idea of solving for an intersection, but with very practical meaning.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Envision Algebra 2 2-7 Additional Practice: Linear-Quadratic Systems Determine the number of solutions for the system of equations: Y = -x^2 + 3x + 2 y = -x^2 + 2x + 18 y = x + 3x – 5 (y = 3x + 2 y = 5x – 10 Y = -x^2 – Zx + […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-223915","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223915","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=223915"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223915\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223915"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223915"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223915"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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