systems of equations multiple choice test doc

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/Meta904 920 0 R 0 G 0.458 0 0 RG /Meta1194 Do /ProcSet[/PDF/Text] 30.699 5.203 TD endobj /Meta520 536 0 R /Meta950 Do /Resources<< >> 174.501 5.203 TD << 1 i Q Q endobj BT q /Font << q Algebra Individual Multiple Choice Test: Unit 6 - Solving Systems of Equations. /Font << ET Q /F3 12.131 Tf q q 1423 0 obj 747 0 obj /F4 24 0 R 0 G endstream /Meta1185 Do /ProcSet[/PDF/Text] Q > > @ = { bjbjBrBr h X X d F ~ ~ ~ ~ Y Y Y dF fF fF fF fF fF fF $ H AK | F Y Y Y Y Y F ~ ~ F S S S Y ~ ~ dF S Y dF S S @ D ~ # o A " PF F 0 F A K ] K D D K D @ Y Y S Y Y Y Y Y F F S Y Y Y F Y Y Y Y K Y Y Y Y Y Y Y Y Y X x : CC MATH I STANDARDS UNIT 3 4.5 SOLVING MULTI-STEP EQUATIONS: PART 1 WARM UP: Solve for the given variable. endobj /F4 12.131 Tf stream q /Matrix [1 0 0 1 0 0] Explanation:- The linear equations given in the problem are, kx - 5y = 2, 6x + 2y = 7 These equations can be written as, kx - 5y - 2 = 0, 6x + 2y - 7 = 0 On comparing it with standard pair of linear equation, we get, a 1 = k, b 1 = -5, c 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that your students need to practice. 0.737 w 1.007 0 0 1.007 45.168 365.866 cm /F4 12.131 Tf /Font << Q 1.502 5.203 TD Q stream q /Type /XObject 0 G /Type /XObject endobj /Type /XObject /Length 68 Q q endobj q Q /Subtype /Form 1434 0 obj >> q << q -0.028 Tc /Font << >> endstream /Font << 0.737 w /Subtype /Image /BBox [0 0 15.59 16.44] >> << Q Q /Subtype /Form Q endobj /Subtype /Form 246 0 obj >> q (-) Tj /Subtype /Form Q Q << Q Q << Q q 0.68 Tc >> q /Meta1518 1536 0 R /FormType 1 /Matrix [1 0 0 1 0 0] endstream /Subtype /Form /Meta1205 Do q Q 1 i /F1 7 0 R q /Font << /Meta756 772 0 R /FormType 1 1 i 0 g /ProcSet[/PDF/Text] /Length 69 404 0 obj -0.486 Tw /Subtype /Form 1 i 1.007 0 0 1.007 130.989 893.588 cm BT 0 g (55) Tj q Q 1 i stream << /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /Meta1031 1047 0 R stream 0 g /F4 24 0 R ET /BBox [0 0 15.59 16.44] endstream 1.007 0 0 1.007 551.058 773.891 cm stream q /Type /XObject /FormType 1 /BBox [0 0 88.214 16.44] /FormType 1 endstream /Meta473 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