Solve using substitution
\(\textbf{1)}\) \(y=-2x+5\)
\(y=x-1\)
The answer is \((2,1)\)
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 \((2, 1)\).
\(\textbf{2)}\) \(y=x+5\)
\(y=2x+4\)
The answer is \((1,6)\)
To solve this system of equations, set the two equations equal to each other:
\(x+5 = 2x+4\)
Subtract \(x\) from both sides:
\(5 = x + 4\)
Subtract 4 from both sides:
\(1 = x\)
So, \(x = 1\)
Now substitute \(x = 1\) into the first equation
\(y = x + 5\):
\(y = 1 + 5\)
\(y = 6\)
Therefore, the solution is \((1, 6)\).
![Link to Youtube Video Solving Question Number 2](https://andymath.com/wp-content/uploads/2020/08/play-video.jpg")
\(\textbf{3)}\) \(4x+3y=12\)
\(y=4\)
The answer is \((0,4)\)
We can substitute \(y = 4\) into the first equation
\(4x + 3y = 12\):
\(4x + 3(4) = 12\)
\(4x + 12 = 12\)
Subtract 12 from both sides:
\(4x = 0\)
Divide both sides by 4:
\(x = 0\)
And we were given y=4 in the initial question
Therefore, the solution is \((0, 4)\).
\(\textbf{4)}\) \(2x-3y=0\)
\(y=x-1\)
The answer is \((3,2)\)
Substitute \(y = x – 1\) into the first equation
\(2x – 3y = 0\):
\(2x – 3(x – 1) = 0\)
Distribute the -3:
\(2x – 3x + 3 = 0\)
Combine like terms:
\(-x + 3 = 0\)
Subtract 3 from both sides:
\(-x = -3\)
Multiply both sides by -1:
\(x = 3\)
Now plug \(x = 3\) into
\(y = x – 1\):
\(y = 3 – 1 = 2\)
The solution is \((3, 2)\).
\(\textbf{5)}\) \(3x+y=10\)
\(x-y=2\)
The answer is \((3,1)\)
Solve the second equation
\(x – y = 2\) for \(x\):
\(x = y + 2\)
Now substitute into the first equation \(3x + y = 10\):
\(3(y + 2) + y = 10\)
Distribute the 3:
\(3y + 6 + y = 10\)
Combine like terms:
\(4y + 6 = 10\)
Subtract 6 from both sides:
\(4y = 4\)
Divide by 4:
\(y = 1\)
Now plug \(y = 1\) into \(x = y + 2\):
\(x = 1 + 2\)
\(x = 3\)
The solution is \((3, 1)\).
\(\textbf{6)}\) \(y=3x+2\)
\(2y-6x=-6\)
The answer is No Solution or “Inconsistent”
Substitute \(y = 3x + 2\) into \(2y – 6x = -6\):
\(2(3x + 2) – 6x = -6\)
Distribute the 2:
\(6x + 4 – 6x = -6\)
Simplify:
\(4 = -6\)
This is a contradiction, which means there is no solution. The lines are parallel and will never intersect.
\(\textbf{7)}\) \(y=-3x+5\)
\(9x+3y=15\)
The answer is Infinitely Many Solutions or “Consistent and Dependent”
We’ll substitute \(y = -3x + 5\) into the second equation \(9x + 3y = 15\):
\(9x + 3(-3x + 5) = 15\)
Distribute:
\(9x – 9x + 15 = 15\)
Simplify:
\(15 = 15\)
This is always true, which means the equations represent the same line. There are infinitely many solutions.
\(\textbf{8)}\) Four times one number added to another number is 20. The second number is 5 more than the first. Find the numbers.
The answer is First number is \(3\). Second number is \(8\)
Let the first number be \(x\) and the second number be \(y\).
From the problem:
\(4x + y = 20\) (Equation 1)
\(y = x + 5\) (Equation 2)
Substitute Equation 2 into Equation 1:
\(4x + (x + 5) = 20\)
Simplify:
\(5x + 5 = 20\)
Subtract 5 from both sides:
\(5x = 15\)
Divide by 5:
\(x = 3\)
Now substitute back to find \(y\):
\(y = 3 + 5 = 8\)
The two numbers are \(3\) and \(8\).
See Related Pages\(\)
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.
