Linear Equations Part 1 Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient.

Linear Equations Part 1 Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient. EXAMPLES : 𝑦=𝑥 2𝑥+5=10 5𝑥−4=2𝑥

Linear Equations Part 1 Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient. EXAMPLES : 𝑦=𝑥 2𝑥+5=10 5𝑥−4=2𝑥 Being able to model real world problems from verbal information into equation form is needed. Look for key words and phrases such as added to, subtracted from, times, etc. to create your equation.

Linear Equations Part 1 Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient. EXAMPLES : 𝑦=𝑥 2𝑥+5=10 5𝑥−4=2𝑥 Being able to model real world problems from verbal information into equation form is needed. Look for key words and phrases such as added to, subtracted from, times, etc. to create your equation. EXAMPLE : A tree was four feet tall when it was planted. It grows 2 feet per year. How many years will it take to grow to a height of 16 feet ?

Linear Equations Part 1 Linear equations model real world relationships and patterns. They are represented by a straight line on a graph. They contain a constant and variables that are sometimes multiplied by a coefficient. EXAMPLES : 𝑦=𝑥 2𝑥+5=10 5𝑥−4=2𝑥 Being able to model real world problems from verbal information into equation form is needed. Look for key words and phrases such as added to, subtracted from, times, etc. to create your equation. EXAMPLE : A tree was four feet tall when it was planted. It grows 2 feet per year. How many years will it take to grow to a height of 16 feet ? 4+2𝑦=16 ** the tree started at 4 feet ( constant of 4 ) ** it grows 2 feet per year ( variable with coefficient of 2 ) ** it grows to 16 feet ( constant of 16 as the result )

Linear Equations Part 1 Our objective for these equations is to solve for the given variable or variables that are not known. There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started.

Linear Equations Part 1 Our objective for these equations is to solve for the given variable or variables that are not known. There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started. EXAMPLE #1 : Solve 2𝑥+12=6 With all linear equations we will : 1. distribute any outside coefficients into parentheses 2. combine any like terms on each side 3. get the variable isolated on one side of the equation

Linear Equations Part 1 Our objective for these equations is to solve for the given variable or variables that are not known. There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started. EXAMPLE #1 : Solve 2𝑥+12=6 With all linear equations we will : 1. distribute any outside coefficients into parentheses 2. combine any like terms on each side 3. get the variable isolated on one side of the equation We do not need to distribute here and terms are already combined on both sides. To isolate the variable, move your constant first AWAY from the variable by doing the opposite operation.

Linear Equations Part 1 Our objective for these equations is to solve for the given variable or variables that are not known. There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started. EXAMPLE #1 : Solve 2𝑥+12=6 With all linear equations we will : 1. distribute any outside coefficients into parentheses 2. combine any like terms on each side 3. get the variable isolated on one side of the equation We do not need to distribute here and terms are already combined on both sides. To isolate the variable, move your constant first AWAY from the variable by doing the opposite operation. 2𝑥+12=6 −12=−12 ** opposite operation 2𝑥=−6

Linear Equations Part 1 Our objective for these equations is to solve for the given variable or variables that are not known. There are specific steps you must follow to solve these equations. Let’s look at a few examples to get you started. EXAMPLE #1 : Solve 2𝑥+12=6 With all linear equations we will : 1. distribute any outside coefficients into parentheses 2. combine any like terms on each side 3. get the variable isolated on one side of the equation We do not need to distribute here and terms are already combined on both sides. To isolate the variable, move your constant first AWAY from the variable by doing the opposite operation. 2𝑥+12=6 −12=−12 ** opposite operation 2𝑥=−6 2𝑥 2 = −6 2 ** again opposite operation 𝑥=−3

Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =17

Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =25 4−6𝑥−3=25 ** distribute any outside coefficients

Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =25 4−6𝑥−3=25 ** distribute any outside coefficients 1−6𝑥=25 ** combine any like terms on both sides

Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =25 4−6𝑥−3=25 ** distribute any outside coefficients 1−6𝑥=25 ** combine any like terms on both sides −1 =−1 ** move constants first −6𝑥=24

Linear Equations Part 1 EXAMPLE #2 : Solve 4−3 2𝑥+1 =25 4−6𝑥−3=25 ** distribute any outside coefficients 1−6𝑥=25 ** combine any like terms on both sides −1 =−1 ** move constants first −6𝑥=24 −6𝑥 −6 = 24 −6 ** divide by any coefficient 𝑥=−4

Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥

Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥 There are no coefficients to distribute or like terms on either side to combine. But variables and constants appear on both sides. Let’s move them to opposite sides. I like my variables on the left side and my constants on the right side.

Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥 −5𝑥= −5𝑥 ** opposite operation 4−7𝑥=−10

Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥 −5𝑥= −5𝑥 ** opposite operation 4−7𝑥=−10 −4 =−4 ** opposite operation −7𝑥=−14

Linear Equations Part 1 EXAMPLE #3 : Solve 4−2𝑥=−10+5𝑥 −5𝑥= −5𝑥 ** opposite operation 4−7𝑥=−10 −4 =−4 ** opposite operation −7𝑥=−14 −7𝑥 −7 = −14 −7 𝑥=2

Linear Equations Part 1 EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the government. 𝑇=0.15 𝑑−7550 +755 Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total dollars Lisa earned during the past year ? a) $15,755 b) $20,550 c) $25,550 d) $30,755

Linear Equations Part 1 EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the government. 𝑇=0.15 𝑑−7550 +755 Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total dollars Lisa earned during the past year ? a) $15,755 b) $20,550 c) $25,550 d) $30,755 Solution : Solve for 𝑑 where 𝑇=3455

Linear Equations Part 1 EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the government. 𝑇=0.15 𝑑−7550 +755 Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total dollars Lisa earned during the past year ? a) $15,755 b) $20,550 c) $25,550 d) $30,755 Solution : Solve for 𝑑 where 𝑇=3455 3455=0.15 𝑑−7550 +755

Linear Equations Part 1 EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the government. 𝑇=0.15 𝑑−7550 +755 Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total dollars Lisa earned during the past year ? a) $15,755 b) $20,550 c) $25,550 d) $30,755 Solution : Solve for 𝑑 where 𝑇=3455 3455=0.15 𝑑−7550 +755 3455=0.15𝑑−1132.50+755 * distribute

Linear Equations Part 1 EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the government. 𝑇=0.15 𝑑−7550 +755 Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total dollars Lisa earned during the past year ? a) $15,755 b) $20,550 c) $25,550 d) $30,755 Solution : Solve for 𝑑 where 𝑇=3455 3455=0.15 𝑑−7550 +755 3455=0.15𝑑−1132.50+755 * distribute 3455=0.15𝑑−377.50 * combine like terms

Linear Equations Part 1 EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the government. 𝑇=0.15 𝑑−7550 +755 Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total dollars Lisa earned during the past year ? a) $15,755 b) $20,550 c) $25,550 d) $30,755 Solution : Solve for 𝑑 where 𝑇=3455 3455=0.15 𝑑−7550 +755 3455=0.15𝑑−1132.50+755 * distribute 3455=0.15𝑑−377.50 * combine like terms +377.50= +377.50 * isolate 𝑑 3832.50=0.15𝑑

Linear Equations Part 1 EXAMPLE #4 : Lisa uses the equation below to estimate the amount of taxes, 𝑇, she owes the government. 𝑇=0.15 𝑑−7550 +755 Lisa estimates that she owes the government $3455 in taxes. What is 𝑑, the total dollars Lisa earned during the past year ? a) $15,755 b) $20,550 c) $25,550 d) $30,755 Solution : Solve for 𝑑 where 𝑇=3455 3455=0.15 𝑑−7550 +755 3455=0.15𝑑−1132.50+755 * distribute 3455=0.15𝑑−377.50 * combine like terms +377.50= +377.50 * isolate 𝑑 3832.50=0.15𝑑 3832.50 0.15 = 0.15𝑑 0.15 𝑑=25550 ANSWER is C