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Review for Unit 1 Test 1. 1 and 2 are a linear pair. 1 = 5x + 25 2 = 4x + 20 Find m2 5x + 25 + 4x + 20 = 180 9x + 45 = 180 9x = 135 x = 15 2 = 4(15) + 20 = 80 Y is=the midpoint of RL x + 4y – 15 + 2x + y – 10 6x – 5y 2. M YI bisects LYE -3x + 10y = 25 x + 4y R – 15 = 2x +Y y – 10 L -x + 3y = 5 2(x + 4y – 15) = 6x – 5y E 2x + 8yI – 30 = 6x – 5y -4x + 13y = 30 mEYL 2(2x + y=–(6x 10)– =5y) 6x – 5y Write of equations to find 4x +and 2ysolve – 20a=system 6x – 5y the values and -2x +of7yx = 20y. x = 25 y = 10 3. Six times an angle’s supplement is 5 less than the angle. Find the measure of the angle’s supplement. 6(180 – x) = x – 5 1080 – 6x = x – 5 1085 = 7x 155 = x Supplement = 180 – 155 = 25 4. 1 and 2 form a right angle. 1 = 10x + 12 2 = 3x Find m1 10x + 12 + 3x = 90 13x + 12 = 90 13x = 78 x=6 1 = 10(6) + 12= 72 5. 1 and 2 are complementary. 1 is 15 more than twice 2 Find m2 2 = x; 1 = 90 – x 90 – x = 2x + 15 75 = 3x 25 = x 6. 1 and 2 are vertical angles. 1 = 3x2 – 2x + 6 2 = x2 – 11x + 2 Find the value of x. 3x2 – 2x + 6 = x2 – 11x + 2 2x2 + 9x + 4 = 0 (2x + 1)(x + 4) = 0 x = -½ x = -4 (both work!) 7. The sum of an angle’s supplement and its complement is equal to 8 times the angle. Find the measure of the angle. 180 – x + 90 – x = 8x 270 – 2x = 8x 270 = 10x 27 = x 8. JM intersects GW at point J GJM = 2x + 5 MJW = x + 25 Draw and label a diagram. W J G M 9. Can you assume: M Y is the midpoint of RL YI bisects LYE R Y E L I d) e) Angle R, Y, L LYI are==collinear a) c)MY SO OU LYE SU isangle right EYI RY YL f)b) =+=YI MYR d) e) yes, you’re BISECTOR c)no! yes no! b/c b) MIDPOINT f)a) e) yes, b/c they aretold vertical angles 10. An angle is twice it’s complement. Find the measure of the angle’s supplement. x = 2(90 – x) x = 180 – 2x 3x = 180 x = 60 Supplement = 180 – 60 = 120 11. Solve the system using either substitution or elimination. 3x + 2y = 11 -6x + y = -32 x=5 y = -2