Can you help me with this problem that i'm stuck on please?
- Thread starter 8dltndus
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mmm4444bot
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Have you seen any examples of the substitution method? (If not, I can find resources for you, like this one at purplemath.)
You forgot to tell us why you're stuck. Is there something specific?
Pick one of the two equations, and solve for y. (y will equal some expression that contains x)
Substitute that expression for y in the other equation.
This yields an equation that contains only x.
Solve that equation.
Use the value that you get for x to find y.
Check your solution by substituting the x and y values into the original two equations.
That's the substitution method.
Please tell us what you've thought about or tried.
You forgot to tell us why you're stuck. Is there something specific?
Pick one of the two equations, and solve for y. (y will equal some expression that contains x)
Substitute that expression for y in the other equation.
This yields an equation that contains only x.
Solve that equation.
Use the value that you get for x to find y.
Check your solution by substituting the x and y values into the original two equations.
That's the substitution method.
Please tell us what you've thought about or tried.
nyc_function
New member
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- Dec 10, 2009
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- 39
System of Linear Equations
3x + y = 5...Equation A
2x - 5y = 9..Equation B
I will solve Equation A for y but you can start with either equation solving one variable in terms of the other.
3x + y = 5
y = 5 - 3x...Here we have y in terms of x.
I will now plug (5 - 3x) for y in Equation B to find my x value.
2x - 5y = 9...This is Equation B.
Let y = (5 - 3x)
2x - 5(5 - 3x) = 9
2x -25 + 15x = 9
17x - 25 = 9
17x = 9 + 25
17x = 34
x = 34/17
x = 2
To find the value of y, plug 2 for x into either Equation A or B (your choice).
I will use Equation A one more time.
3x + y = 5...This is Equation A.
Let x = 2
3(2) + y = 5
6 + y = 5
y = 5 - 6
y = -1
Final answer: x = 2 and y = -1.
This can be written as the point (2,-1), which is the solution to the system of linear equations given. In other words, the point (2,-1) is the location where the two lines meet if you graph each equation on the same xy-plane.
The question says to slove each system usind substitution,and my actual problem is:
3x+y=5
2x-5y=9. how do you solve for x and y in this problem?
please help thanks
3x + y = 5...Equation A
2x - 5y = 9..Equation B
I will solve Equation A for y but you can start with either equation solving one variable in terms of the other.
3x + y = 5
y = 5 - 3x...Here we have y in terms of x.
I will now plug (5 - 3x) for y in Equation B to find my x value.
2x - 5y = 9...This is Equation B.
Let y = (5 - 3x)
2x - 5(5 - 3x) = 9
2x -25 + 15x = 9
17x - 25 = 9
17x = 9 + 25
17x = 34
x = 34/17
x = 2
To find the value of y, plug 2 for x into either Equation A or B (your choice).
I will use Equation A one more time.
3x + y = 5...This is Equation A.
Let x = 2
3(2) + y = 5
6 + y = 5
y = 5 - 6
y = -1
Final answer: x = 2 and y = -1.
This can be written as the point (2,-1), which is the solution to the system of linear equations given. In other words, the point (2,-1) is the location where the two lines meet if you graph each equation on the same xy-plane.
mmm4444bot
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- Oct 6, 2005
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That's sad, :sad: