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Solve using substitution

\(\textbf{1)}\) \(y=-2x+5\)
\(y=x-1\)

Show Answer

The answer is \((2,1)\)

Show Work

To find the solution for the system of equations, we can set the two equations equal to each other and solve for the values of x and y.
\(\,\,\,\,\,-2x+5 = x-1\)

First, let’s isolate the x terms on one side of the equation. We can add 2x to both sides to eliminate the -2x on the left side:

\(5 = x-1 + 2x\)

Now, combine like terms:

\(5 = 3x – 1\)

Next, add 1 to both sides of the equation:

\(5 + 1 = 3x\)

Simplify the left side:

\(6 = 3x\)

Now, divide both sides by 3 to solve for x:

\(\frac{6}{3} = \frac{3x}{3}\)

Simplify:

\(2 = x\)

So, we’ve found that \(x = 2\).

Now that we have the value of x, we can plug it into one of the original equations to solve for y. Let’s use the second equation \(y = x – 1\):

\(y = 2 – 1\)

Simplify:

\(y = 1\)

So, the solution to the system of equations is \((2, 1)\).

\(\textbf{2)}\) \(y=x+5\)
\(y=2x+4\)

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The answer is \((1,6)\)

\(\textbf{3)}\) \(4x+3y=12\)
\(y=4\)

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The answer is \((0,4)\)

\(\textbf{4)}\) \(2x-3y=0\)
\(y=x-1\)

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The answer is \((3,2)\)

\(\textbf{5)}\) \(3x+y=10\)
\(x-y=2\)

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The answer is \((3,1)\)

\(\textbf{6)}\) \(y=3x+2\)
\(2y-6x=-6\)

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The answer is No Solution or “Inconsistent”

\(\textbf{7)}\) \(y=-3x+5\)
\(9x+3y=15\)

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The answer is Infinitely Many Solutions or “Consistent and Dependent”

\(\textbf{8)}\) Four times one number added to another number is 20. The second number is 5 more than the first. Find the numbers.

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The answer is First number is \(3\). Second number is \(8\)

See Related Pages\(\)

\(\bullet\text{ System of Equations Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Wolframalpha.com)}\)

\(\bullet\text{ Solving Systems with Elimination}\)
\(\,\,\,\,\,\,\,\,7x+4y=31 \)
\(\,\,\,\,\,\,\,\,3x+2y=15…\)

\(\bullet\text{ Graphing Systems of Inequalities}\)
\(\,\,\,\,\,\,\,\,y\lt-3x+2 \)
\(\,\,\,\,\,\,\,\,y\ge\frac{1}{2}x-1…\)

\(\bullet\text{ 3 variable systems}\)
\(\,\,\,\,\,\,\,\,2x+3y−5z=−7 \)
\(\,\,\,\,\,\,\,\,3x−6y+4z=3 \)
\(\,\,\,\,\,\,\,\,x+4y+2z=15…\)

\(\bullet\text{ Nonlinear Systems}\)
\(\,\,\,\,\,\,\,\,x^2+y^2=8 \)
\(\,\,\,\,\,\,\,\,y=x…\)

\(\bullet\text{ Andymath Homepage}\)

In Summary

Solving systems of equations with substitution involves using the values from one equation to solve for a variable in the other equation. This is a very common method for solving systems of equations with two or more variables.

Solving systems of equations with substitution is typically covered in a high school or college algebra class.

One real-world example of solving systems of equations with substitution is in finding the dimensions of a rectangular garden given the total area and the perimeter. Another example is in finding the quantity of a product given the total cost and the cost per unit.

The 3 most common methods for solving systems of linear equations is substitution, elimination and graphing.

In a system of equations, a solution is considered dependent if there is more than one solution to the system. This means that the equations are not independent of each other, and one equation can be derived from the other. In this case, the system is said to be dependent because the solutions to the equations are not unique.

An inconsistent system of equations is one that has no solutions. This means that the equations in the system contradict each other and cannot be satisfied simultaneously.

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