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Can someone solve this Linear Systems problem for me?

Can someone solve this Linear Systems problem for me? Topic: Homework due tonight
July 17, 2019 / By Charissa
Question: I hate word problems, its homework and its due tonight and I cannot figure it out. An investor has $45,000 to invest in three types of bonds: short-term, intermediate-term, and long-term. How much should she invest in each type to satisfy the given conditions? Short-term bonds pay 4% annually, intermediate-term bonds pay 5%, and long-term bonds pay 6%. The investor wishes to realize a total annual income of 5.00%, with equal amounts invested in short- and intermediate-term bonds. (Round your answers to the nearest thousand.) Thank you
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Best Answers: Can someone solve this Linear Systems problem for me?

Antonette Antonette | 9 days ago
This is an extremely easy problem because the numbers are very "nice". To get an average return of 5% he should invest 15,000 in each type of account.
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We found more questions related to the topic: Homework due tonight


Antonette Originally Answered: How do you solve systems of linear equations by graphing?
Let's take y=2x-5 y=mx+b b is y-intercept which gives you the clue to graph (0,b) m is slope which gives you the clue that m=rise/run 1. Graph (0,-5). 2. Move up 2 units and 1 unit right. 3. Connect both points. Do the same thing for the rest.
Antonette Originally Answered: How do you solve systems of linear equations by graphing?
note that the coefficient of x is your slope. Rememer y = mx + b; m = slope; b = y-intercept; So with this in mind let's use the example y = 2x - 5 2 is your slop and (-5) is your y - intercept. So at x = 0 => y = 2(0) - 5 => y = -5; So plot the point (0, -5); Then do this with x = 1,2,3, ECT... Then do the same for the next equation and see where they intersect. Doing this without graphing would be setting these two equaling to each other and solve for x.
Antonette Originally Answered: How do you solve systems of linear equations by graphing?
solve x-y=-5 for x: x=y-5 and replace the x in the second equation with y-5 2x+y=-1 2(y-5) + y = -1 simplify, put into slope intercept form, and find your intercepts and slope from there. graph using that info.
Antonette Originally Answered: How do you solve systems of linear equations by graphing?
Make up at least three values of x. Substitute them into the equations to get their corresponding y's. That gives you ordered pairs that you can graph. See where the lines cross.

Antonette Originally Answered: Solve the systems of linear equations using subsitution?
First System: x = -2 y = -4 Second System: x = -8 y = -2 Workings out (in case you wanted them!): First System: y = 5x + 6 y = -2x - 8 Multiply the second equation by 2.5 so that the -2x becomes -5x (which is the negative of the 5x in the first equation). You now have: y = 5x + 6 2.5y = -5x - 20 Add the equations together so that the 5x and -5x cancel out. You now have: 3.5y = -14 Divide both sides of the equation by 3.5. You get: y = -4 Now, substitute y as -4 into one of the equations (I've chosen the first one in the question). -4 = 5x + 6 Minus six from both sides so you're left with only 5x on the right hand side. -10 = 5x Finally, divide both sides of the equation by 5. x = -2 Second System: y = -x - 10 y = x + 6 Add the equations together so that the -x and x cancel out. 2y = -4 Divide both sides of the equation by 2. You get: y = -2 Now, substitute y as -2 into one of the equations (I've chosen the first one in the question again). -2 = -x - 10 Add 10 to both sides so only -x is left on the right hand side. 8 = -x Finally, multiply both sides of the equation by -1. -8 = x ...and that is all! :)
Antonette Originally Answered: Solve the systems of linear equations using subsitution?
11. y=x-3 3x-y=7 ok, so for the reason which you have y=x-3. you be responsive to that y=x-3 soooo you are going to plug that into your different equation 3x-(x-3)=7 now you sparkling up for X as quickly as you detect X, plug that kind into the two equation so which you will sparkling up for Y. i wish this enables

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