PHIL
Okay, so I need to figure out how to solve this problem: "If 2x + 3y = 7 and x - y = 5, what is the value of x?" Alright, let's start by understanding what's being asked here. It looks like a system of linear equations with two variables, x and y. The goal is to find the value of x.
First, I remember that there are a couple of methods to solve such systems: substitution and elimination. I wonder which one would be easier here. Let me write down the equations again to get a clearer picture.
The first equation is 2x + 3y = 7, and the second one is x - y = 5. Hmm. Maybe substitution would work because the second equation can be easily solved for one variable in terms of the other. Let me try that.
Starting with the second equation: x - y = 5. If I solve for x, I can add y to both sides, which gives x = y + 5. That seems straightforward. Now that I have x expressed in terms of y, I can substitute this expression into the first equation wherever there's an x. That way, the first equation will only have y as the variable, which I can solve for. Let's do that.
Substituting x = y + 5 into the first equation: 2(y + 5) + 3y = 7. Let me expand that. 2 times y is 2y, and 2 times 5 is 10. So, the equation becomes 2y + 10 + 3y = 7. Combining like terms, 2y + 3y is 5y, so now it's 5y + 10 = 7.
Next, I need to solve for y. Subtract 10 from both sides: 5y = 7 - 10, which simplifies to 5y = -3. Then, divide both sides by 5: y = -3/5. Okay, so y is -3/5. Now that I have y, I can plug this back into the expression we had for x, which was x = y + 5.
Substituting y = -3/5 into x = y + 5 gives x = (-3/5) + 5. To add these, I need to express 5 as a fraction with denominator 5. So, 5 is 25/5. Then, x = (-3/5) + (25/5) = (22/5). Therefore, x is 22/5.
Wait, let me check if that makes sense. Let me verify the solution by plugging both x and y back into the original equations to ensure they satisfy both.
First, check the second equation: x - y = 5. If x is 22/5 and y is -3/5, then 22/5 - (-3/5) = 22/5 + 3/5 = 25/5 = 5. That works out. Now check the first equation: 2x + 3y = 7. Plugging in x and y: 2*(22/5) + 3*(-3/5) = (44/5) - (9/5) = 35/5 = 7. That also works. So both equations are satisfied, which means the solution is correct.
Alternatively, I could have used the elimination method. Let me try that approach too to confirm. The elimination method involves manipulating the equations so that when we add or subtract them, one variable cancels out. Let's see.
We have:
2x + 3y = 7
x - y = 5
If I want to eliminate one variable, say y, I can multiply the second equation by 3 so that the coefficients of y in both equations are 3 and -3. Let's try that.
https://www.18122.com/url/usdt/usdt0505.JPEG
Multiplying equation 2 by 3: 3*(x - y) = 3*5 => 3x - 3y = 15.
Now, add this new equation to equation 1:
2x + 3y = 7
3x - 3y = 15
(2x + 3x) + (3y - 3y) = 7 + 15
5x + 0 = 22
So, 5x = 22 => x = 22/5. Which is the same result as before. Then, substitute x back into one of the original equations to find y. Let's use equation 2 again: x - y = 5. So, 22/5 - y = 5. Solving for y: y = 22/5 - 5 = 22/5 - 25/5 = -3/5. Same result. So both methods lead to the same answer, which is reassuring.
Another way to check is to graph both equations and see where they intersect. The intersection point should be (22/5, -3/5). But since I can't graph here, I'll stick with the algebraic checks.
Wait, just to make sure I didn't make any arithmetic mistakes. Let me go through substitution again step by step.
From equation 2: x = y + 5.
Substitute into equation 1: 2(y + 5) + 3y = 7.
Expand: 2y + 10 + 3y = 7.
Combine like terms: 5y + 10 = 7.
Subtract 10: 5y = -3.
Divide by 5: y = -3/5. Then x = (-3/5) + 5 = (-3/5) + (25/5) = 22/5. Yep, that's correct.
Alternatively, if I had solved for y in equation 2 instead of x, would that work? Let's see. Equation 2: x - y = 5. Solving for y: y = x - 5. Then substitute into equation 1: 2x + 3(x - 5) = 7. Let's try that.
2x + 3x - 15 = 7 => 5x - 15 = 7 => 5x = 22 => x = 22/5. Same result. So regardless of which variable I solve for first, I end up with the same x value. That's good.
I think I'm confident that x is 22/5. But just to recap, the key steps were:
Choose a method (substitution or elimination). Both work here.
Solve one equation for one variable.
Substitute into the other equation to find the value of that variable.
Back-substitute to find the other variable.
Check the solution in both original equations.
Alternatively, with elimination:
Multiply one or both equations to align coefficients for elimination.
Add/subtract equations to eliminate one variable.
Solve for the remaining variable.
Back-substitute to find the eliminated variable.
Check the solution.
Either way, the answer comes out the same. So, x is 22/5, which is 4.4 in decimal form. But since the question doesn't specify the format, fractional form is probably preferred here.
I don't see any mistakes in the steps, and both methods confirm the same answer. The checks also confirm that the values satisfy both equations. Therefore, I think the correct value of x is 22/5.
To solve the system of equations:
2x + 3y = 7
x - y = 5
Step 1: Solve the second equation for x
From ( x - y = 5 ):
( x = y + 5 ).
Step 2: Substitute ( x = y + 5 ) into the first equation
( 2(y + 5) + 3y = 7 ).
Expand and simplify:
( 2y + 10 + 3y = 7 )
( 5y + 10 = 7 )
Subtract 10:
( 5y = -3 )
Divide by 5:
( y = -\frac{3}{5} ).
Step 3: Substitute ( y = -\frac{3}{5} ) back into ( x = y + 5 )
( x = -\frac{3}{5} + 5 = -\frac{3}{5} + \frac{25}{5} = \frac{22}{5} ).
Verification
Equation 2: ( \frac{22}{5} - \left(-\frac{3}{5}\right) = \frac{25}{5} = 5 ) ✓
Equation 1: ( 2\left(\frac{22}{5}\right) + 3\left(-\frac{3}{5}\right) = \frac{44}{5} - \frac{9}{5} = \frac{35}{5} = 7 ) ✓
Final Answer
The value of ( x ) is (\boxed{\dfrac{22}{5}}).
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