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Write a system of two equations in two unknowns for each problem. Solve each system by the method of your choice.Coffee and doughnuts. On Monday, Archic paid $\$ 3.40$ for three doughnuts and two coffees. On Tuesday he paid $\$ 3.60$ for two doughnuts and three coffees. On Wednesday he was tired of paying the tab and went out for coffee by himself. What was his bill for one doughnut and one coffee?

$$\$ 1.40$$

Algebra

Algebra 2

Chapter 4

Systems of Linear Equations

Section 2

The Addition Method

Polynomials

Systems of Equations and Inequalities

Introduction to Algebra

Graph Linear Functions

Write Linear Equations

Linear Equations and Functions

Matrices and Determinants

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So in this problem, we're told about two trips to it. Sounds like a coffee shop that aren't you made. And so what we're trying to figure out is to cost for one donut and one coffee. So I'm gonna let d represent the cost for a donut, and I'm going to let see represent the cost for coffee. And so what we're gonna do in this type of problem is working a write an equation for each day. So on Monday, we're told that he spent a total of $3.40. So when we add up what he spent on doughnuts and coffee, that's what it should equal. Well, we're told he got three doughnuts, but the cost for a donut is D. So three D will represent the cost for the doughnut. And he got two coffees. So when we add to see to that, that would be the cost for the coffee. So now we've run in an equation for Monday. We're gonna do the same thing for Tuesday. One Tuesday, we're told he spent a total of $3.60 and he bought two doughnuts and three coffees. So two D plus three c will equal $3.60. And now we just have to solve the system of equation. I'm going to do this using the addition. Nothing. I know that's nothing. Cancels right away. So I'm gonna look at my devalues. So I'm going to multiply the first equation by negative two and the second equation by three. So when I multiply that first equation by negative to, I'm gonna get negative 60 minus foresee and negative two times $3.40 would be negative $6.80. Now I'm gonna multiply in the next equation by three. I'm gonna make it red just so we have a little bit of a difference there. Well, three times to D is 60 three times three c is positive nine c and three times $3.60 will give us $10.80. Now we can add the two equations up because negative 60 plus 60 0 cancels and then we have negative foresee plus nine C, which is five C and negative. $6.80 plus $10. 80 cents is just $4. So to get, see we're gonna divide both sides by five now because we're representing money. We want to actually write this as a decimal. And 4/5 is the same as 0.80 in other words, 80 cents. So now we know the cost for coffee. It's 80 cents, but we now we need to cost for the doughnut. So we're gonna pick even the first or second equation. I'll use the first equation, so we have three D plus to see is $3.40 and we're gonna substitute 80 cents in for C. So we'll have three D plus two times 80 cents equals $3.40. Well, two times 80 cents is a dollar 60. So we have three D plus a dollar, 60 equals $3.40. So we're in the subtracted dollar, 60 from both sides, so three D will be equal to a dollar 80 and then we're gonna divide both sides by three. And what we find is we get 60 cents and now we've found the cost for the doughnut is 60 cents, and now we found both the cost of the doughnut and the cost

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