Can anyone help me with my last question for solving problems of linear equations
Sure - what's the question?
Let's call t the number of cans of tomatoes and c the number of cans of corn. We know t + c = 32. We also know that 0.8t + 0.4c = 18. You have two equations now - can you work it from there?
First you set up a system of equations. To show the total amount of veggies bought, you write x+y=32. Then you have to show the cost per can, using .8x+.4y=18. From there, you solve for one variable in each equation, then make the rest of the equation equal to each other, such as y=2x+2 and y=3x goes to 2x+2=3x. Then you can solve for x, and then plug x in to find y.
@Dreams<3 do you still need more help after following matt101's response?
yes in matt 101 response where do i go on from there
let t = 32- c substitute in the 2nd eqn you will have 1 eqn with 1 variable c solve for c now you have c find t using t = 32 - c
am sorry am still a little lost
in the 2nd eqn replace t with 32 - c 0.8(32- c) + 0.4c = 18 distribute, group like terms solve for c
what does it means to subtract by c
\[25.2c=18\] is it going to llok like that
What you're trying to do is solve for y in each equation, so then you can set them equal to each other. When you solve for y (rearrange the equation to get y on its own), you get the equations y=-x+32 and y=-2x+45. You know that y=y, so x+32=-2x+45. When you solve that equation for x, you get x=13. Now, take x and plug it into the original equation. So if x+y=32, then 13+y=32, so you know y=19. Your answer is (13,19).
that should be 25.6
i know but dnt i subtract 0.8 from it
the question ask to find t
when you have something outside the parenthesis you multiply by each term inside
T is tomatoes, and the tomatoes are x. In math, you use x and y for your variables about 85% of the time, so it's natural for me to convert them over. Your x coordinate is the number of tomatoes. When you go back, you can see I used x for t.
0.8 multiply by 32 = 25.6
i got 25.6
0.8(32 - c) + 0.4c = 18 0.8*32 +0.8*-c +0.4c = 18 25.6 -0.8c +0.4c = 18
leave the variables on one side the constants on the other keep the balance of the eqn
is it 1.4
no group like terms let me see what you are doing
ok
Was the answer wrong
do you have c = 19 therefore t = 32 - 19 = 13 remember t = cans of tomatoes check your answer 0.8c + 0.4t = 0.8*13 + 0.4*19= 10.4 + 7.6 = 18.00 as given
0.8(32 - c) + 0.4c = 18 0.8*32 +0.8*-c +0.4c = 18 25.6 -0.8c +0.4c = 18
are you ok with this?
so tha answer is 13
for c yes now find t
t = 32 - c t = 32 - 13 t = 19 do you understand the process ?
i think i kinda do but i have to look at it over again
first write the eqns 2eqns with 2 variables then express one variable in terms of the other example t = 32 - c now substitute in the other eqn to get one eqn with 1 variable solve that eqn to find the value of that variable now use this value for the variable in the eqn to find the value of the other variable once you have the values for the variables check your answer to get started it does not matter which eqn you use first hope this helps
Thank you ....... people have been calling me so much that it made my practice test run out of time on me so now i have to wait till tuesday untill my teacher can rrestart it
this will give you time to review and I hope you will get your priorities in order. Let me know if you need more help. all the best.
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