>> (=) Tj Q q 155 0 obj /Length 167 /Subtype /Form q /Length 343 /ProcSet[/PDF] >> 1.014 0 0 1.007 251.439 506.642 cm 1 i /Subtype /Form BT >> >> >> ET BT /ProcSet[/PDF/Text] 0.045 Tw q /BBox [0 0 534.67 16.44] /F3 17 0 R /Length 67 /Type /XObject /ProcSet[/PDF/Text] 584 0 obj /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] /FormType 1 /Meta336 Do 0.369 Tc >> /Subtype /Form /Font << /Meta577 593 0 R q q stream /Meta1462 Do /BBox [0 0 88.214 16.44] 0.458 0 0 RG >> << /Font << >> /FormType 1 Q /ProcSet[/PDF/Text] 1.007 0 0 1.007 654.946 306.394 cm /ProcSet[/PDF] /Font << 1 i Q 1.005 0 0 1.007 102.382 522.451 cm /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] 1.005 0 0 1.006 102.382 722.699 cm 1 i << Q q /Subtype /Form Q /Resources<< q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Subtype /Form endobj /ProcSet[/PDF/Text] 0.786 Tc /Length 89 >> endobj 0.458 0 0 RG /Font << 0 g 1 i endobj ET Q 0.564 G /ProcSet[/PDF] 1 i 0.564 G q /ProcSet[/PDF] q ET /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] q /Length 54 /Type /XObject << >> Q /Type /XObject /Resources<< >> BT endobj /BBox [0 0 30.642 16.44] 189.467 5.203 TD q Q 0 g BT 1 i 74.891 5.203 TD Q /Type /XObject /Font << << /Meta813 Do /Meta1182 1200 0 R Q Q /Font << /Meta563 Do Q q 0 g /Meta332 Do /ProcSet[/PDF] Q /Meta619 635 0 R q /Meta760 Do /ProcSet[/PDF/Text] Q 1 i [(e)-21(q)-15(ua)-34(ls)] TJ /F3 17 0 R /Meta1097 Do /Type /XObject 1 i q /Length 54 >> Q 5 7 y y < _____7. q /FormType 1 /Meta100 114 0 R >> /Meta591 607 0 R /Meta1033 Do /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] Q ET 1.005 0 0 1.007 102.382 843.902 cm endobj endobj /FormType 1 q /Matrix [1 0 0 1 0 0] endobj Q /BBox [0 0 88.214 16.44] If one line is useful, let's see what we can do with two lines. /Meta1069 1087 0 R 1 i >> q endstream endstream 1650 0 obj 0.737 w stream 0.458 0 0 RG /FormType 1 1.007 0 0 1.007 411.035 183.686 cm 1 i 1 i /Subtype /Form /Meta1383 Do >> /Meta1558 1578 0 R endstream q (A\)) Tj The length is 12ft longer than the width. /Meta892 908 0 R endstream endstream q 1 g q 1 i stream q Q 0 G 0 5.203 TD /Meta1483 Do q 1 i 0 g stream endobj (=) Tj 1 g /Meta480 Do /Meta30 43 0 R BT /BBox [0 0 534.67 16.44] 332 0 obj (=) Tj /BBox [0 0 88.214 16.44] Q (11) Tj q 524 0 obj /Matrix [1 0 0 1 0 0] q endstream /BBox [0 0 88.214 16.44] Q endstream /MaxWidth 2000 (21) Tj 1.007 0 0 1.007 67.753 420.822 cm /Meta634 Do 0 G 0 5.203 TD /Meta1253 Do /Meta1468 1486 0 R >> q 1 i >> 1.007 0 0 1.007 411.035 836.374 cm HOW TO SOLVE MULTI-STEP EQUATIONS 1) SIMPLIFY ALL EXPRESSIONS (Distribute, Combine Like Terms, Evaluate) 2) IDENTIFY VARIABLE 3) CANCEL Operations using their opposites in reverse order of operations EXAMPLES: 1) 7x + 5 - 3x = 17 2) 9(x + 4) = 63 3) 4(2y - 3) = 28 4) 7(6 - x) = 56 5) 3(8+ n) - 15 = 24 6) 7) 19 = 9 + 2(x - 7) 8) /Resources<< <> -0.056 Tw BT /F3 12.131 Tf 25.454 5.203 TD BT 1 i q >> /Subtype /Form /Matrix [1 0 0 1 0 0] /Font << 1105 0 obj q /Resources<< /Resources<< /Meta1475 Do 1 i endstream Q << /Meta1334 1352 0 R [(2)-29(x )] TJ endstream /Matrix [1 0 0 1 0 0] >> 0.737 w 1 i q Q Q 1.007 0 0 1.007 411.035 182.933 cm q q /FormType 1 /Resources<< /Meta1544 Do stream stream 0.564 G Q q /Font << 242 0 obj 0.458 0 0 RG 1 i /Meta269 Do Q /Type /XObject q 1200 0 obj /ProcSet[/PDF/Text] /FormType 1 q Q endobj 0 g Q 0 20.154 m 1 i /Type /XObject >> q /ProcSet[/PDF/Text] 0 G q >> 866 0 obj ET /Resources<< 0 G << 267 0 obj /ProcSet[/PDF/Text] /F4 24 0 R >> /Resources<< 1.007 0 0 1.007 551.058 350.81 cm stream 0.564 G /Font << q endobj 0 G Q /Subtype /Form /F3 17 0 R BT 1 i /Matrix [1 0 0 1 0 0] >> /Type /XObject BT endstream q >> Q q /ProcSet[/PDF/Text] stream >> /Type /XObject q q Q /FormType 1 1.007 0 0 1.007 551.058 457.709 cm q /Resources<< >> 0 g Q 1.007 0 0 1.007 130.989 432.867 cm /Subtype /Form /Meta296 Do /F4 24 0 R Worked example: equivalent systems of equations, Worked example: non-equivalent systems of equations, Level up on the above skills and collect up to 480 Mastery points, Systems of equations number of solutions: fruit prices (1 of 2), Systems of equations number of solutions: fruit prices (2 of 2), Solutions to systems of equations: consistent vs. inconsistent, Solutions to systems of equations: dependent vs. independent, Number of solutions to a system of equations, Number of solutions to a system of equations graphically, Number of solutions to a system of equations algebraically. q >> 1 i q 41.187 20.154 l /Meta1216 1234 0 R /Subtype /Form 1 i >> << 1625 0 obj 1 i Q (=) Tj q Find 2A +3B. /F3 12.131 Tf q /Resources<< /FormType 1 35.942 5.336 TD /Meta343 357 0 R /F3 12.131 Tf 1.007 0 0 1.007 411.035 457.709 cm /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Meta1693 1713 0 R /Font << endobj << /Subtype /Form 0 g 1186 0 obj /ProcSet[/PDF] /BBox [0 0 534.67 16.44] /Length 16 >> /FormType 1 (a) (b) ( c) (d) (e) 3. endobj /FormType 1 Q /Length 16 /BBox [0 0 88.214 16.44] Q Q /Matrix [1 0 0 1 0 0] /Length 69 /F2 12.131 Tf /Matrix [1 0 0 1 0 0] Q /F4 24 0 R /BBox [0 0 88.214 35.886] /Meta326 340 0 R 1157 0 obj 0 g /Resources<< /FormType 1 /F3 12.131 Tf (\() Tj /ProcSet[/PDF/Text] endobj /Type /XObject Q Q >> q /Subtype /Form /Meta1166 Do 1 i endobj /Matrix [1 0 0 1 0 0] endobj (A\)) Tj Q /Meta1072 Do /Meta1290 Do /Meta1452 Do Q /ProcSet[/PDF/Text] /Subtype /Form Q endstream /Type /XObject /Resources<< /Font << /F3 17 0 R BT q stream 252 0 obj >> >> /FormType 1 >> /Resources<< BT >> /FormType 1 /Meta484 500 0 R 0 g endstream (A\)) Tj /ProcSet[/PDF/Text] /Subtype /Form Q 0 G /Meta1469 Do 1167 0 obj /Font << /BBox [0 0 534.67 16.44] 764 0 obj BT endstream endobj /Matrix [1 0 0 1 0 0] /FormType 1 /Length 16 1.502 5.203 TD q /ProcSet[/PDF] 0 g /Length 12 stream /Matrix [1 0 0 1 0 0] Students can use this quiz and worksheet to practice the following skills: Reading comprehension - ensure that you draw the most important information from the related lesson on the system of . /FormType 1 506 0 obj 0 5.203 TD /Length 59 /F3 12.131 Tf Q 1.007 0 0 1.006 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TANNEBERG - ALGEBRA 1 Chapter 5 - Systems of LInear Equations Chapter 5 Calendar, Overview, and Vocab 5-1 SOLVING SYSTEMS OF EQUATIONS BY GRAPHING (Monday, January 8) Notes: 5-1_solving_by_graphing_notes.pdf Assignment: 5-1_solving_by_graphing_worksheet.pdf 1204 0 obj Q /Meta324 Do /ProcSet[/PDF/Text] stream 1.014 0 0 1.007 531.485 836.374 cm /Resources<< /Type /XObject 709 0 obj 0.369 Tc 0 w /Resources<< q 0.564 G /BBox [0 0 88.214 16.44] q /FormType 1 679.036 420.822 m /F3 12.131 Tf Q /Meta121 135 0 R Q 0 5.203 TD /Matrix [1 0 0 1 0 0] endobj /Meta314 328 0 R Q q endstream 0 g 820 0 obj 0.458 0 0 RG q >> 0 G << Q /Meta35 Do Q ____ 1. /BBox [0 0 30.642 16.44] /Meta1576 1596 0 R >> /FormType 1 /Length 167 0 G 0 g 1.005 0 0 1.006 102.382 843.902 cm endobj /F3 12.131 Tf 0 G /Meta1213 Do q 1.014 0 0 1.007 111.416 527.721 cm stream endobj /F4 24 0 R >> /Length 16 /Meta717 Do 1.007 0 0 1.007 45.168 680.542 cm /Type /XObject /Resources<< endstream Q /BBox [0 0 30.642 16.44] 0.369 Tc 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