The sum of two numbers is 112. The second is 7 more than 4 times the first. What are the two numbers?
Am I doing something wrong? I seem to be stuck.
x + y =112
(x+7)4y=112
The second equation should be:
y+7=4x
Should be the 2nd equation.
The 2nd one:
y
is 7 more. This is a little tricky. "Is" means an equals sign, but you're also saying that number to left left is 7 more than what's on the right. So you still have to add 7 to the left to balance it out.
y+7=
"four times the first" is:
4x
y+7 = 4x
Can you solve it from there?
Matt
Made a mistake. In my paragraph that reads "is 7 more" I should say SUBTRACT 7 from the left side:
y-7=
That will leave you with the equation:
y-7=4x
My apologies.
Matt
No problem, Matt! It's important to correct any mistakes to get the correct answer. Now, let's solve the equations you have correctly identified:
1. x + y = 112
2. y - 7 = 4x
To solve these equations, we can use the method of substitution or elimination. I'll explain how to solve them using the substitution method:
From equation 1, we can isolate x:
x = 112 - y
Now, substitute this value of x into equation 2:
y - 7 = 4(112 - y)
Now, distribute 4:
y - 7 = 448 - 4y
Next, combine like terms by adding 4y to both sides of the equation:
5y - 7 = 448
To isolate y, add 7 to both sides of the equation:
5y = 455
Finally, divide both sides by 5 to solve for y:
y = 455/5
Simplifying, we find:
y = 91
Now, substitute this value of y back into equation 1 to find x:
x + 91 = 112
Subtract 91 from both sides:
x = 21
So, the two numbers are x = 21 and y = 91.