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Mathematics 12 Online
OpenStudy (anonymous):

solve by substitution method 0.05x-0.01y=2.6 0.26x+0.18y=22.8

OpenStudy (triciaal):

usually easier to work with whole numbers so start by getting rid of the fraction multiply by 100

OpenStudy (anonymous):

5x-y=26 26x+18y=228

OpenStudy (triciaal):

to solve by substitution express one variable in terms of the other then use in the second equation. you will have one equation with one variable. solve for this variable. use this value in the original to find the value of the other variable.

OpenStudy (anonymous):

i dont understand how to do that

OpenStudy (triciaal):

using the first equation 5x - y = 260 y = 5x -260 using this in the second equation 26 x + 18 (5x - 260) = 2280 can you continue

OpenStudy (anonymous):

116x-4680=2280

OpenStudy (triciaal):

keep going

OpenStudy (anonymous):

x=60

OpenStudy (triciaal):

you have x now find y

OpenStudy (anonymous):

by pluggin in 60 to 5x-y=260?

OpenStudy (triciaal):

yes

OpenStudy (anonymous):

y=-40

OpenStudy (triciaal):

watch your sign

OpenStudy (anonymous):

its a negative isnt it?

OpenStudy (triciaal):

y = 5x -260 = 5*60 -260 =300- 260 = 40 y = 40

OpenStudy (anonymous):

5x60+y=260 300+y=260 subtract -300 y=-40

OpenStudy (triciaal):

5x-y = 260

OpenStudy (anonymous):

which is correct

OpenStudy (triciaal):

look at the original

OpenStudy (anonymous):

ok so we have to swtich it and it is a postive

OpenStudy (anonymous):

the correct answer is 40

OpenStudy (triciaal):

x = 60 and y = 40 always good practice to check your solution in the original

OpenStudy (anonymous):

ok thanks

OpenStudy (triciaal):

plug these values in the second equation to check

OpenStudy (triciaal):

here is a review of the steps 1. pick a variable and find the expression for that variable using one of the equations 2. use this expression in the other equation. 3. solve for only variable in this equation 4. use this value to find the value of the other variable. 5. check your solution in either of the original equation any questions?

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