2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations Name Date Student ID 1. Section Suppose a well-designed survey shows that orangeis the favorite color for 4out of every 9 students at a particular school. If there are 585students at the school, what would be the expected number of students at the school whose favorite color is orange? students Solution We will begin by letting x =the number of students at the school whose favorite color is orange. Then, we will translate the problem into a proportion, taking care that the units are in the same location on each ratio. 4 like orange x like orange = 9 students 585 students Next, we will multiply both sides of the equation by the LCD, 585students and solve for xas follows. 585 students ⋅ 4 like orange 9 students = 585 students ⋅ x like orange 585 students 65 (4 like orange) = 1 (x like orange) 260 = x Therefore, we could expect approximately 260students to say their favorite color is orange. https://homework.derivita.com/static/printassignment.html 1/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 2. Solve for n. 7 13 = 2 4 n− n= Solution To get nby itself, we use the Addition Property of Equality. Add 7 to each side to ‘undo’ the 2 subtraction, then simplify. 7 2 7 + 2 n − 13 4 7 + 2 = 27 4 n = We can check our work by plugging 27 in for nin the original equation: 4 27 7 − 4 2 27 14 − 4 4 13 4 Since we have a true statement, we know n = https://homework.derivita.com/static/printassignment.html ? = ? = ✓ = 13 4 13 4 13 4 27 is a valid solution. 4 2/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 3. Solve for x. x + 4.9 = −8.25 x= Solution To get xby itself, we use the Subtraction Property of Equality. Subtract 4.9from each side to ‘undo’ the addition, then simplify. x +4.9 = −8.25 −4.9 −4.9 x = −13.15 We can check our work by plugging −13.15in for xin the original equation: ? −13.15 + 4.9 = −8.25 ✓ −8.25 = −8.25 Since we have a true statement, we know x = −13.15is a valid solution. https://homework.derivita.com/static/printassignment.html 3/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 4. Use the phrase below to answer the following questions. Sixlessthan mis equal to 28. Translate the phrase into an algebraic equation. Equation Now, use your equation to solve for m. m= Solution The phrase “less than” tells us that we are subtracting 6from m, and the phrase “is equal to” indicates an ‘=’ sign. Therefore, the algebraic equation is m − 6 = 28 To solve for m, add 6to both sides. m −6 = +6 m = 28 +6 34 Be sure to plug this value into the original equation to check that it is a valid solution. https://homework.derivita.com/static/printassignment.html 4/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 5. Solve for n. −n = 26 n= Solution Recall that −nis equivalent to −1n. To isolate n on one side of the equation we use the Division Property of Equality. Divide both sides of the equation by −1. −n = 26 −1n = 26 −1n −1 = 26 −1 n = −26 Check this answer by plugging it back in to the original equation. ? −(−26) = 26 ✓ 26 = 26 https://homework.derivita.com/static/printassignment.html 5/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 6. Solve for x. 6 x = 18 7 x= Solution To isolate xon one side of the equation we use the Multiplication Property of Equality. Multiply both 6 sides of the equation by the reciprocal of . 7 6 x = 18 7 7 6 7 ⋅ x = (18) 6 7 6 7 18 1x = ⋅ 6 1 x = 21 Check this answer by plugging it back in to the original equation. 6 ? (21) = 18 7 ? 6(3) = 18 ✓ 18 = 18 https://homework.derivita.com/static/printassignment.html 6/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 7. Solve for y . 7 + 12y + 6 − 11y = 25 y= Solution To solve for y , we first need to simplify each side. Start by rearranging the terms and then combining like terms. 7 + 12y + 6 − 11y = 25 12y − 11y + 7 + 6 = 25 y + 13 = 25 Now that both sides are fully simplified, we can isolate y by subtracting 13from both sides. y + 13 −13 = −13 y 25 = 12 We can check our work by plugging 12in for y in the original equation: ? 7 + 12 (12) + 6 − 11 (12) = 25 ? 7 + 144 + 6 − 132 = 25 ✓ 25 = 25 Since we have a true statement, we know y = 12is a valid solution. https://homework.derivita.com/static/printassignment.html 7/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 8. Solve for x. −3 (5x + 4) + 16 (x − 3) = −2x + 2 (x − 3) x= Solution To solve for x, we first need to simplify each side. Start by distributing, then rearrange the terms and combine like terms. −3 (5x + 4) + 16 (x − 3) = −2x + 2 (x − 3) −15x − 12 + 16x − 48 = −2x + 2x − 6 −15x + 16x − 12 − 48 = −6 x − 60 = −6 Now that both sides are fully simplified, we can isolate xby adding 60to both sides. x − 60 +60 = +60 = x −6 54 We can check our work by plugging 54in for xin the original equation: ? −3 (5 (54) + 4) + 16 ((54) − 3) = −2 (54) + 2 ((54) − 3) ? −3 (274) + 16 (51) = −108 + 2 (51) ? −822 + 816 = −108 + 102 ✓ −6 = −6 Since we have a true statement, we know x = 54is a valid solution. https://homework.derivita.com/static/printassignment.html 8/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 9. Solve for n. −4 (n − 6) − 3 = 3 n= Solution Begin by simplifying each side of the equation, using the distributive property first. −4 (n − 6) − 3 = 3 −4n − 4 (−6) − 3 = 3 −4n + 21 = 3 Continue to isolate non one side of the equation. −4n + 21 −21 = 3 −21 −4n = −18 −4n −4 = −18 −4 = 9 2 n Be sure to check this answer by plugging it back in to the original equation. https://homework.derivita.com/static/printassignment.html 9/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 10. Solve for x. −9x = −x + 16 x= Solution In this equation, we have variables on both sides. We need to make one side the variable side and the other the constant side. Since there is already a constant on the right, let's make the variable side be on the left. To achieve this, we need to add xto both sides of the equation. Doing so, we get −9x = −x + 16 −9x + x = −x + x + 16 −8x = 16 Dividing both sides by −8and simplifying, we get −8x = 16 −8x −8 = 16 −8 x = −2 We can check this solution by plugging it into the original equation. ? −9 (−2) = − (−2) + 16 ✓ 18 = 18 Therefore, x = −2is a valid solution. https://homework.derivita.com/static/printassignment.html 10/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 11. Solve for y . 8y + 8 = 12y − 6 y= Solution This equation has variables and constants on each side. Since 12 > 8, let's make the variable side be on the right and the constant side be on the left. Doing so, we get 8y + 8 = 12y − 6 8y − 8y + 8 = 12y − 8y − 6 8 = 4y − 6 8 + 6 = 4y − 6 + 6 14 = 4y Dividing both sides by 4and simplifying, we get 14 = 4y 14 4 7 2 = 4y 4 =y We can check this solution by sunstituting it into the original equation. 8⋅ 7 7 ? + 8 = 12 ⋅ −6 2 2 ? 28 + 8 = 42 − 6 ✓ 36 = 36 Therefore, y = 7 is a valid solution. 2 https://homework.derivita.com/static/printassignment.html 11/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 12. Solve the following equation. If there are multiple answers, enter each answer separated by a comma. Enter Noneif there are no solutions. ∣13x + 5∣ = 6 x= Solution This equation will be true any time the expression inside the absolute value bars is equal to 6or −6. This leads us to the following two equations, 13x + 5 = 6or 13x + 5 = −6. Solving the first equation, we get, 13x + 5 = 6 13x + 5 − 5 = 6 − 5 13x = 1 13x 13 1 = 13 1 13 x = Then, solving the second equation, we get 13x + 5 = −6 13x + 5 − 5 = −6 − 5 13x = −11 13x 13 Therefore the solution to ∣13x + 5∣ = 6is x = https://homework.derivita.com/static/printassignment.html =− 11 13 x =− 11 13 1 11 or − . 13 13 12/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 13. Solve the following equation. If there are multiple answers, enter each answer separated by a comma. Enter Noneif there are no solutions. ∣13x + 5∣ = −6 x= Solution In general, any equation written in the form ∣ax + b∣ = cwhere c < 0will have no solutions. Since the given equation is an absolute value expression equal to a negative number, we can conclude there are no solutions. https://homework.derivita.com/static/printassignment.html 13/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 14. Given the equation −12x + 3y = 9 a) Solve for y if x = −1. y= b) Solve for y in general. y= Solution a) To solve for y when x = −1, plug the given value in for x, then isolate y . −12x + 3y = 9 −12 (−1) + 3y = 9 12 + 3y = 9 12 − 12 + 3y = 9 − 12 3y = −3 3 3 y =− 3 3 y = −1 ) To solve for y in general, we follow the same process as above, but we work with variables instead of b plugging in any values. −12x + 3y = 9 −12x + 12x + 3y = 9 + 12x 3y = 9 − (−12x) 3 9 − (−12x) y = 3 3 y = 3 + 4x https://homework.derivita.com/static/printassignment.html 14/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 15. Solve for z . x + y + z + 10 = v z= Solution To isolate z , we need to subtract the other terms from both sides of the equation. x + y + z + 10 = v x − x + y − y + z + 10 − 10 = v − x − y − 10 z = v − x − y − 10 16. Solve for x. y = 4x + 9xz x= Solution To solve the given equation for x, we start by rewriting the right-hand side by using the distributive property and then solving. Doing so, we get y = 4x + 9xz y = x (4 + 9z ) y 4 + 9z y 4 + 9z Therefore, the rewritten equation is x = https://homework.derivita.com/static/printassignment.html = x (4 + 9z ) 4 + 9z =x y . 4 + 9z 15/16 2/3/22, 12:42 PM Derivita: SAT PREP Diagnostic test: Linear Equations SAT PREP Diagnostic test: Linear Equations 17. Solve for x. 4 (x + 3) + 8y = −16 x= Solution To isolate x, we need to distribute, then we can subtractterms fromboth sides of the equation. 4 (x + 3) + 8y = −16 4x + 12 + 8y = −16 4x + 12 + 8y − 12 − 8y = −16 − 12 − 8y 4x = −28 − 8y Finally, we can divide both sides by 4. 4x = −28 − 8y 4x 4 = −28 − 8y 4 x =− 28 8y − 4 4 x = −7 − 2y https://homework.derivita.com/static/printassignment.html 16/16