lesson 6 homework practice solve simultaneous linear equations answer key

Worksheet on Simultaneous Linear Equations

Practice each pair of the equation problems from the worksheet on simultaneous linear equations with the two variables and two linear equations. Solving simultaneous linear equations with two variables by using the method of substitution to solve each pair of the equations and also solve the equations by using the method of elimination.

1.  Use the method of substitution to solve each other of the pair of simultaneous equations:    (a) x + y = 15                     x - y = 3

  (b) x + y = 0                       x - y = 2

  (c) 2x - y = 3                    4x + y = 3

  (d) 2x - 9y = 9               5x + 2y = 27

  (e) x + 4y = -4               3y - 5x = -1

  (f) 2x - 3y = 2                  x + 2y = 8

  (g) x + y = 7                  2x - 3y = 9

  (h) 11y + 15x = -23        7y - 2x = 20

   (i) 5x - 6y = 2                6x - 5y = 9

2. Solve each other pair of equation given below using elimination method:   (a) x + 2y = -4                        3x - 5y = -1

  (b) 4x + 9y = 5                      -5x + 3y = 8

  (c) 9x - 6y = 12                      4x + 6y = 14

  (d) 2y - (3/x) = 12                   5y + (7/x) = 1

  (e) (3/x) + (2/y) = (9/xy)       (9/x) + (4/y) = (21/xy)

  (f) (4/y) + (3/x) = 8             (6/y) + (5/x) = 13

  (g) 5x + (4/y) = 7                    4x + (3/y) = 5

  (h) x + y = 3                         -3x + 2y = 1

  (i) -3x + 2y = 5                       4x + 5y = 2

3. Solve the following simultaneous equations:   (a) 3a + 4b = 43                      -2a + 3b = 11

  (b) 4x - 3y = 23                      3x + 4y = 11

  (c) 5x + (4/y) = 7                    4x + (3/y) = 5

  (d) 4/(p - 3) + 6/(q - 4) = 5       5/(p - 3) - 3/(q - 4) = 1

  (e) (l/6) - (m/15) = 4               (l/3) - (m/12) = 19/4

  (f) 3x + 2y = 8                        4x + y = 9

  (g) x - y = -1                         2y + 3x = 12

  (h) (3y/2) - (5x/3) = -2           (y/3) + (x/3) = 13/6

  (i) x - y = 3                           (x/3) + (y/2) = 6

  (j) (2x/3) + (y/2) = -1             (-x/3) + y = 3

  (k) 5x + 8y = 9                       2x + 3y = 4

  (l) 3 - 2(3a - 4b) = -59            (a - 3)/4 - (b - 4)/5 = 2¹/₁₀

Answers for the worksheet on simultaneous linear equations are given below to check the exact answers of the above questions on system of linear equations.

1. (a) x = 9, y = 6

(b) x = 1, y = -1

(c) x = 1, y = -1

(d) x = 261/49, y = 9/49

(e) x = -8/23, y = -21/23

(f) x = 4, y = 2

(g) x = 6, y = 1

(h) x = -3, y = 2

(i) x = 4, y = 3

2. (a) x = -2, y = -1

(b) x = -1, y = 1

(c) x = 2, y = 1

(d) x = -1/2, y = 3

(e) x = 3, y = 1

(f) x = 1/2, y = 2

(g) x = -1, y = 1/3

(h) x = 1, y = 2

(i) x = -21/23, y = 26/23

3. (a) a = 5, b = 7

(b) x = 5, y = -1

(c) x = -1, y = 1/3

(d) p = 5, q = 6

(e) l = -2, m = -65

(f) x = 2, y = 1

(g) x = 2, y = 3

(h) x = 141/38, y = 53/19

(i) x = 9, y = 6

(j) x = -3, y = 2

(k) x = 5, y = -2

(l) a = 5, b = -4

●   Simultaneous Linear Equations

Simultaneous Linear Equations

Comparison Method

Elimination Method

Substitution Method

Cross-Multiplication Method

Solvability of Linear Simultaneous Equations

Pairs of Equations

Word Problems on Simultaneous Linear Equations

Practice Test on Word Problems Involving Simultaneous Linear Equations

●   Simultaneous Linear Equations - Worksheets

Worksheet on Problems on Simultaneous Linear Equations

8th Grade Math Practice From Worksheet on Simultaneous Linear Equations to HOME PAGE

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4.2: Solve Systems of Equations by Substitution

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Learning Objectives

By the end of this section, you will be able to:

• Solve a system of equations by substitution
• Solve applications of systems of equations by substitution

Before you get started, take this readiness quiz.

• Simplify −5(3−x). If you missed this problem, review Example 1.10.43 .
• Simplify 4−2(n+5). If you missed this problem, review Example 1.10.49 .
• Solve for y. 8y−8=32−2y If you missed this problem, review Example 2.3.22 .
• Solve for x. 3x−9y=−3 If you missed this problem, review Example 2.6.22 .

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

\(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Example \(\PageIndex{1}\).

Example \(\PageIndex{1}\): How to Solve a System of Equations by Substitution

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

Try It \(\PageIndex{2}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{-2 x+y=-11} \\ {x+3 y=9}\end{array}\right.\)

Try It \(\PageIndex{3}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+3 y=10} \\ {4 x+y=18}\end{array}\right.\)

SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.

• Solve one of the equations for either variable.
• Substitute the expression from Step 1 into the other equation.
• Solve the resulting equation.
• Substitute the solution in Step 3 into one of the original equations to find the other variable.
• Write the solution as an ordered pair.
• Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Example \(\PageIndex{4}\).

Example \(\PageIndex{4}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=-1} \\ {y=x+5}\end{array}\right.\)

The second equation is already solved for y . We will substitute the expression in place of y in the first equation.

Try It \(\PageIndex{5}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x+y=6} \\ {y=3 x-2}\end{array}\right.\)

Try It \(\PageIndex{6}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-y=1} \\ {y=-3 x-6}\end{array}\right.\)

(−1,−3)

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y .

Example \(\PageIndex{7}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{3 x+y=5} \\ {2 x+4 y=-10}\end{array}\right.\)

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Try It \(\PageIndex{8}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x+y=2} \\ {3 x+2 y=-1}\end{array}\right.\)

(1,−2)

Try It \(\PageIndex{9}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{-x+y=4} \\ {4 x-y=2}\end{array}\right.\)

In Example \(\PageIndex{7}\) it was easiest to solve for y in the first equation because it had a coefficient of 1. In Example \(\PageIndex{10}\) it will be easier to solve for x .

Example \(\PageIndex{10}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-2 y=-2} \\ {3 x+2 y=34}\end{array}\right.\)

We will solve the first equation for \(x\) and then substitute the expression into the second equation.

Try It \(\PageIndex{11}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-5 y=13} \\ {4 x-3 y=1}\end{array}\right.\)

(−2,−3)

Try It \(\PageIndex{12}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-6 y=-6} \\ {2 x-4 y=4}\end{array}\right.\)

When both equations are already solved for the same variable, it is easy to substitute!

Example \(\PageIndex{13}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=-2 x+5} \\ {y=\frac{1}{2} x}\end{array}\right.\)

Since both equations are solved for y , we can substitute one into the other.

Try It \(\PageIndex{14}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=3 x-16} \\ {y=\frac{1}{3} x}\end{array}\right.\)

Try It \(\PageIndex{15}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{y=-x+10} \\ {y=\frac{1}{4} x}\end{array}\right.\)

Be very careful with the signs in the next example.

Example \(\PageIndex{16}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x+2 y=4} \\ {6 x-y=8}\end{array}\right.\)

We need to solve one equation for one variable. We will solve the first equation for y .

Try It \(\PageIndex{17}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{x-4 y=-4} \\ {-3 x+4 y=0}\end{array}\right.\)

\((2,\frac{3}{2})\)

Try It \(\PageIndex{18}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-y=0} \\ {2 x-3 y=5}\end{array}\right.\)

\((−\frac{1}{2},−2)\)

In Example , it will take a little more work to solve one equation for x or y .

Example \(\PageIndex{19}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{4 x-3 y=6} \\ {15 y-20 x=-30}\end{array}\right.\)

We need to solve one equation for one variable. We will solve the first equation for x .

Try It \(\PageIndex{20}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{2 x-3 y=12} \\ {-12 y+8 x=48}\end{array}\right.\)

infinitely many solutions

Try It \(\PageIndex{21}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x+2 y=12} \\ {-4 y-10 x=-24}\end{array}\right.\)

Look back at the equations in Example \(\PageIndex{22}\). Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Example \(\PageIndex{22}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-2 y=-10} \\ {y=\frac{5}{2} x}\end{array}\right.\)

The second equation is already solved for y , so we can substitute for y in the first equation.

Try It \(\PageIndex{23}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{3 x+2 y=9} \\ {y=-\frac{3}{2} x+1}\end{array}\right.\)

no solution

Try It \(\PageIndex{24}\)

Solve the system by substitution. \(\left\{\begin{array}{l}{5 x-3 y=2} \\ {y=\frac{5}{3} x-4}\end{array}\right.\)

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

HOW TO USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.

• Read the problem. Make sure all the words and ideas are understood.
• Identify what we are looking for.
• Name what we are looking for. Choose variables to represent those quantities.
• Translate into a system of equations.
• Solve the system of equations using good algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Example \(\PageIndex{25}\)

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Try It \(\PageIndex{26}\)

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

The numbers are 3 and 7.

Try It \(\PageIndex{27}\)

The sum of two number is −6. One number is 10 less than the other. Find the numbers.

The numbers are 2 and −8.

In the Example \(\PageIndex{28}\), we’ll use the formula for the perimeter of a rectangle, P = 2 L + 2 W .

Example \(\PageIndex{28}\)

The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and width of the rectangle.

Try It \(\PageIndex{29}\)

The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.

The length is 12 and the width is 8.

Try It \(\PageIndex{30}\)

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.

The length is 23 and the width is 6.

For Example \(\PageIndex{31}\) we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

Example \(\PageIndex{31}\)

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Try It \(\PageIndex{32}\)

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of the angles are 22 degrees and 68 degrees.

Try It \(\PageIndex{33}\)

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of the angles are 36 degrees and 54 degrees.

Example \(\PageIndex{34}\)

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

Try It \(\PageIndex{35}\)

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

There would need to be 160 policies sold to make the total pay the same.

Try It \(\PageIndex{36}\)

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Kenneth would need to sell 1,000 suits.

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

• Instructional Video-Solve Linear Systems by Substitution
• Instructional Video-Solve by Substitution

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Simultaneous Linear Equations Worksheets

There are times when two linear equations share an ordered pair. Under those circumstances you can use them in coordination with one another to find a common solution. A system is a group of linear equations. When a system contains equations that all share an ordered pair, we call them simultaneous. There are a number of different ways that we can use to find the ordered pair that solve the system. You will see them listed below. What that means is that when all of these lines are graphed they will intersect at that one single point. These worksheets and lessons will help us better understand how to approach and ultimately solve simultaneous linear equations.

Aligned Standard: Grade 8 Expressions and Equations - 8.EE.C.7b

• Solve the System Step-by-Step Lesson - Solve the system in each case. This is a very basic starting point for students.
• Guided Lesson - Graph the systems to find the solution for the systems.
• Guided Lesson Explanation - These are very straightforward questions for you to begin to tackle with students.
• Independent Practice - Ten problems to practice away with. Get your graph paper ready.
• Matching Worksheet - Match the systems with their graphs.
• Answer Keys - These are for all the unlocked materials above.

Homework Sheets

I start this skill in a very basic manner. A flat line is always a good place to start.

• Homework 1 - Solve this system of equations by graphing. Graph the equations, and then write the solution.
• Homework 2 - Calculate the solution by graphing.
• Homework 3 - This equation tells you that every y-value is 8. Plot some points that have a y value of 8, like (0, 8) and (1, 8), and then draw a line connecting them.

Practice Worksheets

"Solving systems" always makes it sound a lot more complicated than the topic really is.

• Practice 1 - See where they all meet up?
• Practice 2 - Take it all home for you.
• Practice 3 - Make the Ys equal to each other and get this one going.

Math Skill Quizzes

For the answers, I tried to give you a snapshot of where the graph intersection is found.

• Quiz 1 - Number three and four are a cinch.
• Quiz 2 - Don't let the fractions through.
• Quiz 3 - How much does the line rise and run?

What are Simultaneous Linear Equations?

Linear equations, a relatively newer concept for you as you get into middle school, is a significant tool required to use in everyday applications. These equations are very helpful in describing a relationship between two variables in the physical world. They allow scientists to make predictions and conversions and calculate rates. Graphing linear equations make the trends that you can see in the data it produces visible. For example: y = 2x + 1. This is a linear equation.

When two linear equations contain the same variables (x and y) we can use them together to solve for these variables this is called simultaneous linear equations. Example of such two equations is: 2x - 3y = 4, 3x + y = 1. The solution a simultaneous equation is presented in the form of an ordered pair (x, y). That solution will satisfy both equations meaning that they are points that can be found on each of the equations lines when they are graphed on coordinate system.

How to Solve Simultaneous Linear Equations

There are 3 basic methods for determining an ordered pair that would satisfy both linear equations:

The steps to solve simultaneous linear equations are as follows:

Elimination Method

Step # 1: Multiply each equation with a number that makes the leading co-efficient the same.

Step # 2: Find the difference between the two-equations. Subtract second from the first.

Step # 3: Now solve the new equation that has only one variable.

Step # 4: Substitute the value of this variable in either of the equations and find the other.

Step # 5: Write the answer in an ordered pair.

I would say this is a helpful method but does not always work perfectly if the lead coefficients are very large or odd values.

Substitution Method

This is where you rearrange one of the equations to satisfy a single variable. For example, you make one of equations equal to x. You then plug (substitute) this value in for the x of the other equation. You then solve the equation algebraically to find an ordered pair that solves the system. The great thing about this is that you can check your answer by plugging them into both equations and see if it does in fact work.

Graphing Method

If you graph both equations, wherever they intersect is the solution we are looking for. That ordered is the ordered pair we are looking for.

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Multi-Step Equations Exercises

Multi-step equations practice problems with answers.

For this exercise, I have prepared seven (7) multi-step equations for you to practice. If you feel the need to review the techniques involved in solving multi-step equations, take a short detour to review my other lesson about it. Click the link below to take you there!

Solving Multi-Step Equations

1) Solve the multi-step equation for [latex]\large{c}[/latex].

[latex]c – 20 = 4 – 3c[/latex]

Add both sides by [latex]20[/latex]. Next, add [latex]3c[/latex] to both sides. Finally, divide both sides by the coefficient of [latex]4c[/latex] which is [latex]4[/latex] to get [latex]c=6[/latex].

2) Solve the multi-step equation for [latex]\large{n}[/latex].

[latex] – \,4\left( { – 3n – 8} \right) = 10n + 20[/latex]

• Remember to always perform the same operation on both sides of the equation.
• Subtract by [latex]32[/latex].
• Subtract by [latex]10n[/latex].
• Divide by [latex]2[/latex]
• The final solution is [latex]n=-6[/latex].

3) Solve the multi-step equation for [latex]\large{y}[/latex].

[latex]2\left( {4 – y} \right) – 3\left( {y + 3} \right) = – 11[/latex]

Apply twice the Distributive Property of Multiplication over Addition to the left side of the equation. Then combine like terms . Add both sides by [latex]1[/latex] followed by dividing both sides of the equation by [latex]-5[/latex].

4) Solve the multi-step equation for [latex]\large{k}[/latex].

[latex]{\Large{{6k + 4} \over 2}} = 2k – 11[/latex]

Multiply both sides by [latex]2[/latex]. Next, subtract [latex]4[/latex] to both sides. Then, subtract [latex]4k[/latex]. Finally, divide by [latex]2[/latex] to obtain the value of [latex]k[/latex] which is [latex]-13[/latex].

5) Solve the multi-step equation for [latex]\large{x}[/latex].

[latex] – \left( { – 8 – 3x} \right) = – 2\left( {1 – x} \right) + 6x[/latex]

Apply the Distributive Property on both sides of the equation. Be careful when multiplying expressions with the same or different signs . Next, add [latex]2[/latex] to both sides, then subtract [latex]3x[/latex], and finally finish it off by dividing [latex]5[/latex] to both sides.

6) Solve the multi-step equation for [latex]\large{m}[/latex].

[latex]{\large{3 \over 4}}m – 2\left( {m – 1} \right) = {\large{1 \over 4}}m + 5[/latex]

7) Solve the multi-step equation for [latex]\large{x}[/latex].

[latex]3\left( {3x – 8} \right) – 5\left( {3x – 8} \right) = 4\left( {x – 2} \right) – 6\left( {x – 2} \right)[/latex]

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Word Problems on Simultaneous Linear Equations | Simultaneous Linear Equations Questions

By solving the system of linear equations in two variables, you will get an ordered pair having x coordinate and y coordinate values (x, y) that satisfies both equations. Here those simultaneous linear equations are in the form of word problems. So, you can solve different word problems with the help of linear equations.

We have already learned some steps and methods to solve the simultaneous linear equations in two variables. Assume the unknown quantities in the question as x, y variables and represent them in the form of a linear equation according to the condition mentioned in the question. And follow the methods to solve the formed system of linear equations to get the values of unknown quantities. We have also provided simultaneous equations problems with solutions that help you to grasp the concept.

Simultaneous Equations Word Problems

One number is greater than thrice the other number by 6. If 4 times the smaller number exceeds the greater by 7, find the numbers?

Let x, y be the two numbers

So that x > y

Given that, one number is greater than thrice the other number by 6.

Then, we can write it as

x = 3y + 6 ——— (i)

According to the question,

4 times the smaller number exceeds the greater by 7.

4y – x = 7 ——— (ii)

Substitute (3y + 6) in equation (ii).

4y – (3y + 6) = 7

4y – 3y – 6 = 7

y – 6 = 7

Substitute y = 13 in equation (i)

x = 3(13) + 6

So, the two numbers are 45, 13.

The sum of two numbers is 25 and their difference is 5. Find those two numbers?

Let the two numbers be x and y.

The sum of two numbers is 25

x + y = 25 —— (i)

The difference of numbers is 5.

x – y = 5 —– (ii)

Add both the equations (i) & (ii)

x + y + x – y = 25 + 5

Putting x = 15 in equation (i)

15 + y = 25

y = 25 – 15

So, the two numbers are 15, 10.

The sum of two numbers is 50. If the larger is doubled and the smaller is tripled, the difference is 25. Find the two numbers.

Let the two numbers be x and y

According to the given question,

The sum of two numbers is 50.

x + y = 50 —— (i)

The larger number is doubled and the smaller number is tripled, the difference is 25.

2x – 3y = 25 —— (ii)

Multiply the first equation by 3.

3(x + y) = 3 x 50

3x + 3y = 150 —— (iii)

Add equation (ii) and equation (iii)

2x – 3y + 3x + 3y = 25 + 150

Substituting the value of x in equation (i)

35 + y = 50

y = 50 – 35

Hence the two numbers are 35, 15.

The class IX students of a certain public school wanted to give a farewell party to the outgoing students of class X. They decided to purchase two kinds of sweets, one costing Rs. 70 per kg and the other costing Rs. 85 per kg. They estimated that 34 kg of sweets were needed. If the total money spent on sweets was Rs. 2500, find how much sweets of each kind they purchased?

Let the quantity of sweets purchased be x kg which cost Rs. 70 per kg and sweets purchased y kg which cost Rs. 85 per kg.

x + y = 34 ——- (i)

70x + 85y = 2500 ——– (ii)

Multiplying the equation (i) by 70, we get

70(x + y) = 34 x 70

70x + 70y = 2380 —– (iii)

Subtracting equation (iii) from equation (ii), we get

70x + 70y – (70x + 85y) = 2380 – 2500

70x + 70y – 70x – 85y = -120

-15y = -120

Substitute y = 1.6 in equation (i)

x + 1.6 = 34

x = 34 – 1.6

Hence, the sweets purchased 1 kg 600 grams which cost Rs. 85 per kg and 32 kg 400 grams which cost Rs. 70 per Kg.

A two-digit number is eight times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number?

Let the two-digit number be xy.

Given that, the two digit number is eight times the sum of its digits.

xy = 8(x + y) —– (i)

In the two digit number xy, x is in the tens position and y is in ones position.

xy = 10 . x + 1 . y

xy = 10x + y ——- (ii)

Substitute equation (ii) in equation (i)

10x + y = 8(x + y)

10x + y = 8x + 8y

10x – 8x = 8y – y

2x – 7y = 0 —– (iii)

The number formed by reversing the digits 18 less than the given number.

xy – yx = 18

(10 . x + 1 . y) – (10 . y + 1 . x) = 18

(10x + y) – (10y + x) = 18

10x + y – 10y – x = 18

9x – 9y = 18

9(x – y) = 18

(x – y) = 18/9

x – y = 2 —— (iv)

Multiply equation (iv) by 2.

2(x – y) = 2 x 2

2x – 2y = 4 —— (v)

Subtract equation (v) from equation (iii)

(2x – 2y) – (2x – 7y) = 4 – 0

2x – 2y – 2x + 7y = 4

Substitute y = 4/5 in equation (iv)

x – 4/5 = 2

x = 2 + 4/5

x = (10 + 4)/5

So, the required two digit number is 144/5.

If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to ⅗ and if the numerator and denominator are each diminished by 1, the fraction becomes equal to ⅔, find the fraction.

Let the fraction be x/y.

According the given question,

(x + 2) / (y + 1) = ⅗

5(x + 2) = 3(y + 1)

5x + 10 = 3y + 3

5x – 3y = 3 – 10

5x – 3y = -7 —— (i)

(x – 1) / (y – 1) = ⅔

3(x – 1) = 2(y – 1)

3x – 3 = 2y – 2

3x – 2y = -2 + 3

3x – 2y = 1 —— (ii)

Multiply the equation (i) by 2, equation (ii) by 3.

2(5x – 3y) = 2 x -7 ⇒ 10x – 6y = -14

3(3x – 2y) = 3 x 1 ⇒ 9x – 6y = 3

(10x – 6y) – (9x – 6y) = -14 – 3

10x – 6y – 9x + 6y = -17

Substitute x = -17 in equation (i)

5(-17) – 3y = -7

-85 – 3y = -7

-3y = -7 + 85

y = -78 / 3

Therefore, the fraction is 17/26.

If four times the age of the son is added to the age of the father, the sum is 64. But if twice the age of the father is added to the age of the son, the sum is 82. Find the ages of father and son?

Let the father’s age be x years, the son’s age be y years.

4y + x = 64 —– (i)

2x + y = 82 —— (ii)

Multiply the equation (i) by 2

2(x + 4y) = 64 x 2

2x + 8y = 128

Subtract 2x + 8y = 128 from equation (ii)

(2x + 8y) – (2x + y) = 128 – 82

2x + 8y – 2x – y = 46

y = 6 years 7 months

Substituting y = 46/7 in equation (ii)

2x + 46/7 = 82

2x = 82 – 46/7

2x = (574 – 64)/7

x = 36 years 4 months

Therefore, the father’s age is 36 years 4 months, the son’s age is 6 years 7 months.

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Systems of Linear Equations Worksheets and Answer Keys

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Unit 1: Algebra foundations

Unit 2: solving equations & inequalities, unit 3: working with units, unit 4: linear equations & graphs, unit 5: forms of linear equations, unit 6: systems of equations, unit 7: inequalities (systems & graphs), unit 8: functions, unit 9: sequences, unit 10: absolute value & piecewise functions, unit 11: exponents & radicals, unit 12: exponential growth & decay, unit 13: quadratics: multiplying & factoring, unit 14: quadratic functions & equations, unit 15: irrational numbers, unit 16: creativity in algebra.

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Unit 6 – Linear Equations and Inequalities

Solutions to Equations

LESSON/HOMEWORK

LECCIÓN/TAREA

LESSON VIDEO

EDITABLE LESSON

EDITABLE KEY

SMART NOTEBOOK

Two-Step Equations

More Work with Two-Step Equations

Manipulating Expressions within Equations

Recognizing Structure to Solve Two-Step Equations

Solving Word Problems with Two-Step Equations Day 1

Solving Word Problems with Two-Step Equations Day 2

Solving Two-Step Inequalities

An Interesting Property of Inequalities

Modeling with Inequalities

Unit Review

Unit 6 Review

UNIT REVIEW

REPASO DE LA UNIDAD

EDITABLE REVIEW

Unit 6 Assessment Form A

EDITABLE ASSESSMENT

Unit 6 Assessment – Form B

Unit 6 Exit Tickets

Unit 6 Mid-Unit Quiz – Form A

Unit 6 Mid-Unit Quiz – Form B

U06.AO.01 – Practice Solving Two Step Equations

EDITABLE RESOURCE

U06.AO.02 – Practice with Modeling with Two-Step Equations

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